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Does a member have to be initialized to take its address?
Is it safe to use the “this” pointer in an initialization list?C++ Get class member address in constructor initialization listWhat does the explicit keyword mean?How to initialize private static members in C++?How to initialize all members of an array to the same value?Efficiency of Java “Double Brace Initialization”?Initialization of an ArrayList in one lineStatic constant string (class member)How do C++ class members get initialized if I don't do it explicitly?JavaScript check if variable exists (is defined/initialized)C++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?How to directly initialize a HashMap (in a literal way)?
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Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
#include <string>
class Klass
public:
Klass()
: ptr_str&str
, str
private:
std::string *ptr_str;
std::string str;
;
this question is similar to mine, but the order is correct there, and the answer says
I'd advise against coding like this in case someone changes the order of the members in your class.
Which seems to mean reversing the order would be illegal but I couldn't be sure.
c++ pointers initialization language-lawyer
add a comment |
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
#include <string>
class Klass
public:
Klass()
: ptr_str&str
, str
private:
std::string *ptr_str;
std::string str;
;
this question is similar to mine, but the order is correct there, and the answer says
I'd advise against coding like this in case someone changes the order of the members in your class.
Which seems to mean reversing the order would be illegal but I couldn't be sure.
c++ pointers initialization language-lawyer
Just curious why do you need it?
– vahancho
May 10 at 13:45
1
@vahancho I am adding a private vector to the end of a class (so it is physically separate from the interface) but this vector takes pointers to some public members. I would like to know if I have to move it up or if this is fine.
– Ayxan
May 10 at 13:51
1
just make sure you do not try to access the member before its actually there
– formerlyknownas_463035818
May 10 at 13:56
2
@Ayxan Language-lawyer questions might bring into this site interesting discussions. I would therefore suggest to wait a little longer with such a quick acceptance of the first answer. Though it may be correct, other opinions might be valuable as well.
– Daniel Langr
May 10 at 14:01
@DanielLangr nice suggestion.
– Ayxan
May 10 at 14:03
add a comment |
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
#include <string>
class Klass
public:
Klass()
: ptr_str&str
, str
private:
std::string *ptr_str;
std::string str;
;
this question is similar to mine, but the order is correct there, and the answer says
I'd advise against coding like this in case someone changes the order of the members in your class.
Which seems to mean reversing the order would be illegal but I couldn't be sure.
c++ pointers initialization language-lawyer
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
#include <string>
class Klass
public:
Klass()
: ptr_str&str
, str
private:
std::string *ptr_str;
std::string str;
;
this question is similar to mine, but the order is correct there, and the answer says
I'd advise against coding like this in case someone changes the order of the members in your class.
Which seems to mean reversing the order would be illegal but I couldn't be sure.
c++ pointers initialization language-lawyer
c++ pointers initialization language-lawyer
edited May 10 at 14:16
Ayxan
asked May 10 at 13:43
AyxanAyxan
2,722724
2,722724
Just curious why do you need it?
– vahancho
May 10 at 13:45
1
@vahancho I am adding a private vector to the end of a class (so it is physically separate from the interface) but this vector takes pointers to some public members. I would like to know if I have to move it up or if this is fine.
– Ayxan
May 10 at 13:51
1
just make sure you do not try to access the member before its actually there
– formerlyknownas_463035818
May 10 at 13:56
2
@Ayxan Language-lawyer questions might bring into this site interesting discussions. I would therefore suggest to wait a little longer with such a quick acceptance of the first answer. Though it may be correct, other opinions might be valuable as well.
– Daniel Langr
May 10 at 14:01
@DanielLangr nice suggestion.
– Ayxan
May 10 at 14:03
add a comment |
Just curious why do you need it?
– vahancho
May 10 at 13:45
1
@vahancho I am adding a private vector to the end of a class (so it is physically separate from the interface) but this vector takes pointers to some public members. I would like to know if I have to move it up or if this is fine.
– Ayxan
May 10 at 13:51
1
just make sure you do not try to access the member before its actually there
– formerlyknownas_463035818
May 10 at 13:56
2
@Ayxan Language-lawyer questions might bring into this site interesting discussions. I would therefore suggest to wait a little longer with such a quick acceptance of the first answer. Though it may be correct, other opinions might be valuable as well.
– Daniel Langr
May 10 at 14:01
@DanielLangr nice suggestion.
– Ayxan
May 10 at 14:03
Just curious why do you need it?
– vahancho
May 10 at 13:45
Just curious why do you need it?
– vahancho
May 10 at 13:45
1
1
@vahancho I am adding a private vector to the end of a class (so it is physically separate from the interface) but this vector takes pointers to some public members. I would like to know if I have to move it up or if this is fine.
– Ayxan
May 10 at 13:51
@vahancho I am adding a private vector to the end of a class (so it is physically separate from the interface) but this vector takes pointers to some public members. I would like to know if I have to move it up or if this is fine.
– Ayxan
May 10 at 13:51
1
1
just make sure you do not try to access the member before its actually there
– formerlyknownas_463035818
May 10 at 13:56
just make sure you do not try to access the member before its actually there
– formerlyknownas_463035818
May 10 at 13:56
2
2
@Ayxan Language-lawyer questions might bring into this site interesting discussions. I would therefore suggest to wait a little longer with such a quick acceptance of the first answer. Though it may be correct, other opinions might be valuable as well.
– Daniel Langr
May 10 at 14:01
@Ayxan Language-lawyer questions might bring into this site interesting discussions. I would therefore suggest to wait a little longer with such a quick acceptance of the first answer. Though it may be correct, other opinions might be valuable as well.
– Daniel Langr
May 10 at 14:01
@DanielLangr nice suggestion.
– Ayxan
May 10 at 14:03
@DanielLangr nice suggestion.
– Ayxan
May 10 at 14:03
add a comment |
2 Answers
2
active
oldest
votes
Does a member have to be initialized to take its address?
No.
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
Yes. Yes.
There is no restriction that operand of unary & need to be initialised. There is an example in the standard in specification of unary & operator:
int a;
int* p1 = &a;
Here, the value of a
is uninitialized and it is OK to point to it.
What that example doesn't demonstrate is pointing to an object before its lifetime has begun, which is what happens in your example. Using a pointer to an object before and after its lifetime is explicitly allowed if the storage is occupied. Standard draft says:
[basic.life] Before the lifetime of an object has started but after the storage which the object will occupy has been allocated or, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, any pointer that represents the address of the storage location where the object will be or was located may be used but only in limited ways ...
The rule goes on to list how the usage is restricted. You can get by with common sense. In short, you can treat it as you could treat a void*
, except violating these restrictions is UB rather than ill-formed. Similar rule exists for references.
There are also restrictions on computing the address of non-static members specifically. Standard draft says:
[class.cdtor] ... To form a pointer to (or access the value of) a direct non-static member of an object
obj
, the construction ofobj
shall have started and its destruction shall not have completed, otherwise the computation of the pointer value (or accessing the member value) results in undefined behavior.
In the constructor of Klass
, the construction of Klass
has started and destruction hasn't completed, so the above rule is satisfied.
P.S. Your class is copyable, but the copy will have a pointer to the member of another instance. Consider whether that makes sense for your class. If not, you will need to implement custom copy and move constructors and assignment operators. A self-reference like this is a rare case where you may need custom definitions for those, but not a custom destructor, so it is an exception to the rule of five (or three).
P.P.S If your intention is to point to one of the members, and no object other than a member, then you might want to use a pointer to member instead of pointer to object.
7
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
1
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
1
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variablea
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.
– Brian
May 10 at 14:57
1
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
add a comment |
Funny question.
It is legitimate and will "work", though barely. There is a little "but" related to types which makes the whole thing a bit awkward with a bad taste (but not illegitimate), and which might make it illegal some border cases involving inheritance.
You can, of course, take the address of any object whether it's initialized or not, as long as it exists in the scope and has a name which you can prepend operator&
to. Dereferencing the pointer is a different thing, but that wasn't the question.
Now, the subtle problem is that the standard defines the result of operator&
for non-static struct members as "“pointer to member of class C of type T” and is a prvalue designating C::m".
Which basically means that ptr_str&str
will take the address of str
, but the type is not pointer-to, but pointer-to-member-of. It is then implicitly and silently cast to pointer-to.
In other words, although you do not need to explicitly write &this->str
, that's nevertheless what its type is -- it's what it is and what it means [1].
Is this
valid, and is it safe to use it within the initializer list? Well yes, just... barely. It's safe to use it as long as it's not being used to access uninitialized members or virtual functions, directly or indirectly. Which, as it happens, is the case here (it might not be the case in a different, arguably contrieved case).
[1] Funnily, paragraph 4 starts with a clause that says that no member pointer is formed when you put stuff in parenthese. That's remarkable because most people would probably do that just to be 100% sure they got operator precedence right. But if I read correctly, then
&this->foo
and &(this->foo)
are not in any way the same!
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it usesthis
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?
– Ayxan
May 10 at 17:21
1
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not usingthis
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).
– Damon
May 10 at 17:32
2
No,&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write&Klass::str
. Similarly,&this->foo
and&(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.
– Ben Voigt
May 10 at 17:50
2
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
1
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
add a comment |
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2 Answers
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active
oldest
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oldest
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active
oldest
votes
Does a member have to be initialized to take its address?
No.
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
Yes. Yes.
There is no restriction that operand of unary & need to be initialised. There is an example in the standard in specification of unary & operator:
int a;
int* p1 = &a;
Here, the value of a
is uninitialized and it is OK to point to it.
What that example doesn't demonstrate is pointing to an object before its lifetime has begun, which is what happens in your example. Using a pointer to an object before and after its lifetime is explicitly allowed if the storage is occupied. Standard draft says:
[basic.life] Before the lifetime of an object has started but after the storage which the object will occupy has been allocated or, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, any pointer that represents the address of the storage location where the object will be or was located may be used but only in limited ways ...
The rule goes on to list how the usage is restricted. You can get by with common sense. In short, you can treat it as you could treat a void*
, except violating these restrictions is UB rather than ill-formed. Similar rule exists for references.
There are also restrictions on computing the address of non-static members specifically. Standard draft says:
[class.cdtor] ... To form a pointer to (or access the value of) a direct non-static member of an object
obj
, the construction ofobj
shall have started and its destruction shall not have completed, otherwise the computation of the pointer value (or accessing the member value) results in undefined behavior.
In the constructor of Klass
, the construction of Klass
has started and destruction hasn't completed, so the above rule is satisfied.
P.S. Your class is copyable, but the copy will have a pointer to the member of another instance. Consider whether that makes sense for your class. If not, you will need to implement custom copy and move constructors and assignment operators. A self-reference like this is a rare case where you may need custom definitions for those, but not a custom destructor, so it is an exception to the rule of five (or three).
P.P.S If your intention is to point to one of the members, and no object other than a member, then you might want to use a pointer to member instead of pointer to object.
7
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
1
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
1
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variablea
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.
– Brian
May 10 at 14:57
1
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
add a comment |
Does a member have to be initialized to take its address?
No.
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
Yes. Yes.
There is no restriction that operand of unary & need to be initialised. There is an example in the standard in specification of unary & operator:
int a;
int* p1 = &a;
Here, the value of a
is uninitialized and it is OK to point to it.
What that example doesn't demonstrate is pointing to an object before its lifetime has begun, which is what happens in your example. Using a pointer to an object before and after its lifetime is explicitly allowed if the storage is occupied. Standard draft says:
[basic.life] Before the lifetime of an object has started but after the storage which the object will occupy has been allocated or, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, any pointer that represents the address of the storage location where the object will be or was located may be used but only in limited ways ...
The rule goes on to list how the usage is restricted. You can get by with common sense. In short, you can treat it as you could treat a void*
, except violating these restrictions is UB rather than ill-formed. Similar rule exists for references.
There are also restrictions on computing the address of non-static members specifically. Standard draft says:
[class.cdtor] ... To form a pointer to (or access the value of) a direct non-static member of an object
obj
, the construction ofobj
shall have started and its destruction shall not have completed, otherwise the computation of the pointer value (or accessing the member value) results in undefined behavior.
In the constructor of Klass
, the construction of Klass
has started and destruction hasn't completed, so the above rule is satisfied.
P.S. Your class is copyable, but the copy will have a pointer to the member of another instance. Consider whether that makes sense for your class. If not, you will need to implement custom copy and move constructors and assignment operators. A self-reference like this is a rare case where you may need custom definitions for those, but not a custom destructor, so it is an exception to the rule of five (or three).
P.P.S If your intention is to point to one of the members, and no object other than a member, then you might want to use a pointer to member instead of pointer to object.
7
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
1
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
1
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variablea
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.
– Brian
May 10 at 14:57
1
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
add a comment |
Does a member have to be initialized to take its address?
No.
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
Yes. Yes.
There is no restriction that operand of unary & need to be initialised. There is an example in the standard in specification of unary & operator:
int a;
int* p1 = &a;
Here, the value of a
is uninitialized and it is OK to point to it.
What that example doesn't demonstrate is pointing to an object before its lifetime has begun, which is what happens in your example. Using a pointer to an object before and after its lifetime is explicitly allowed if the storage is occupied. Standard draft says:
[basic.life] Before the lifetime of an object has started but after the storage which the object will occupy has been allocated or, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, any pointer that represents the address of the storage location where the object will be or was located may be used but only in limited ways ...
The rule goes on to list how the usage is restricted. You can get by with common sense. In short, you can treat it as you could treat a void*
, except violating these restrictions is UB rather than ill-formed. Similar rule exists for references.
There are also restrictions on computing the address of non-static members specifically. Standard draft says:
[class.cdtor] ... To form a pointer to (or access the value of) a direct non-static member of an object
obj
, the construction ofobj
shall have started and its destruction shall not have completed, otherwise the computation of the pointer value (or accessing the member value) results in undefined behavior.
In the constructor of Klass
, the construction of Klass
has started and destruction hasn't completed, so the above rule is satisfied.
P.S. Your class is copyable, but the copy will have a pointer to the member of another instance. Consider whether that makes sense for your class. If not, you will need to implement custom copy and move constructors and assignment operators. A self-reference like this is a rare case where you may need custom definitions for those, but not a custom destructor, so it is an exception to the rule of five (or three).
P.P.S If your intention is to point to one of the members, and no object other than a member, then you might want to use a pointer to member instead of pointer to object.
Does a member have to be initialized to take its address?
No.
Can I initialize a pointer to a data member before initializing the member? In other words, is this valid C++?
Yes. Yes.
There is no restriction that operand of unary & need to be initialised. There is an example in the standard in specification of unary & operator:
int a;
int* p1 = &a;
Here, the value of a
is uninitialized and it is OK to point to it.
What that example doesn't demonstrate is pointing to an object before its lifetime has begun, which is what happens in your example. Using a pointer to an object before and after its lifetime is explicitly allowed if the storage is occupied. Standard draft says:
[basic.life] Before the lifetime of an object has started but after the storage which the object will occupy has been allocated or, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, any pointer that represents the address of the storage location where the object will be or was located may be used but only in limited ways ...
The rule goes on to list how the usage is restricted. You can get by with common sense. In short, you can treat it as you could treat a void*
, except violating these restrictions is UB rather than ill-formed. Similar rule exists for references.
There are also restrictions on computing the address of non-static members specifically. Standard draft says:
[class.cdtor] ... To form a pointer to (or access the value of) a direct non-static member of an object
obj
, the construction ofobj
shall have started and its destruction shall not have completed, otherwise the computation of the pointer value (or accessing the member value) results in undefined behavior.
In the constructor of Klass
, the construction of Klass
has started and destruction hasn't completed, so the above rule is satisfied.
P.S. Your class is copyable, but the copy will have a pointer to the member of another instance. Consider whether that makes sense for your class. If not, you will need to implement custom copy and move constructors and assignment operators. A self-reference like this is a rare case where you may need custom definitions for those, but not a custom destructor, so it is an exception to the rule of five (or three).
P.P.S If your intention is to point to one of the members, and no object other than a member, then you might want to use a pointer to member instead of pointer to object.
edited May 10 at 15:15
answered May 10 at 13:55
eerorikaeerorika
92k667138
92k667138
7
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
1
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
1
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variablea
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.
– Brian
May 10 at 14:57
1
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
add a comment |
7
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
1
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
1
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variablea
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.
– Brian
May 10 at 14:57
1
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
7
7
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
It would be nice to quote relevant parts of the Standard. These might be relevant: eel.is/c++draft/basic.life#6.sentence-1 and eel.is/c++draft/class.cdtor#3.sentence-2.
– Daniel Langr
May 10 at 13:58
1
1
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
@DanielLangr Thanks for looking those up.
– eerorika
May 10 at 14:05
1
1
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variable
a
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.– Brian
May 10 at 14:57
There's a subtle difference between your example and the OP's example. In your example, the lifetime of the variable
a
has begun, but it was not initialized. In the OP's example, the address of a member is taken before its lifetime. Given that your quotations of the standard already adequately address the OP's question, I think the example should be deleted.– Brian
May 10 at 14:57
1
1
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
@Brian OP's specifically asks about initialization. The quoted rule doesn't explicitly address initialization (no rule addresses it; there simply isn't a rule related to initialization to my knowledge). Sure, lack of lifetime implies lack of initialization, but I prefer to be explicit. In my opinion, the example demonstrating lack of initialization is useful even if it is technically made redundant by the lifetime rule.
– eerorika
May 10 at 15:10
add a comment |
Funny question.
It is legitimate and will "work", though barely. There is a little "but" related to types which makes the whole thing a bit awkward with a bad taste (but not illegitimate), and which might make it illegal some border cases involving inheritance.
You can, of course, take the address of any object whether it's initialized or not, as long as it exists in the scope and has a name which you can prepend operator&
to. Dereferencing the pointer is a different thing, but that wasn't the question.
Now, the subtle problem is that the standard defines the result of operator&
for non-static struct members as "“pointer to member of class C of type T” and is a prvalue designating C::m".
Which basically means that ptr_str&str
will take the address of str
, but the type is not pointer-to, but pointer-to-member-of. It is then implicitly and silently cast to pointer-to.
In other words, although you do not need to explicitly write &this->str
, that's nevertheless what its type is -- it's what it is and what it means [1].
Is this
valid, and is it safe to use it within the initializer list? Well yes, just... barely. It's safe to use it as long as it's not being used to access uninitialized members or virtual functions, directly or indirectly. Which, as it happens, is the case here (it might not be the case in a different, arguably contrieved case).
[1] Funnily, paragraph 4 starts with a clause that says that no member pointer is formed when you put stuff in parenthese. That's remarkable because most people would probably do that just to be 100% sure they got operator precedence right. But if I read correctly, then
&this->foo
and &(this->foo)
are not in any way the same!
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it usesthis
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?
– Ayxan
May 10 at 17:21
1
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not usingthis
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).
– Damon
May 10 at 17:32
2
No,&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write&Klass::str
. Similarly,&this->foo
and&(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.
– Ben Voigt
May 10 at 17:50
2
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
1
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
add a comment |
Funny question.
It is legitimate and will "work", though barely. There is a little "but" related to types which makes the whole thing a bit awkward with a bad taste (but not illegitimate), and which might make it illegal some border cases involving inheritance.
You can, of course, take the address of any object whether it's initialized or not, as long as it exists in the scope and has a name which you can prepend operator&
to. Dereferencing the pointer is a different thing, but that wasn't the question.
Now, the subtle problem is that the standard defines the result of operator&
for non-static struct members as "“pointer to member of class C of type T” and is a prvalue designating C::m".
Which basically means that ptr_str&str
will take the address of str
, but the type is not pointer-to, but pointer-to-member-of. It is then implicitly and silently cast to pointer-to.
In other words, although you do not need to explicitly write &this->str
, that's nevertheless what its type is -- it's what it is and what it means [1].
Is this
valid, and is it safe to use it within the initializer list? Well yes, just... barely. It's safe to use it as long as it's not being used to access uninitialized members or virtual functions, directly or indirectly. Which, as it happens, is the case here (it might not be the case in a different, arguably contrieved case).
[1] Funnily, paragraph 4 starts with a clause that says that no member pointer is formed when you put stuff in parenthese. That's remarkable because most people would probably do that just to be 100% sure they got operator precedence right. But if I read correctly, then
&this->foo
and &(this->foo)
are not in any way the same!
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it usesthis
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?
– Ayxan
May 10 at 17:21
1
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not usingthis
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).
– Damon
May 10 at 17:32
2
No,&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write&Klass::str
. Similarly,&this->foo
and&(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.
– Ben Voigt
May 10 at 17:50
2
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
1
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
add a comment |
Funny question.
It is legitimate and will "work", though barely. There is a little "but" related to types which makes the whole thing a bit awkward with a bad taste (but not illegitimate), and which might make it illegal some border cases involving inheritance.
You can, of course, take the address of any object whether it's initialized or not, as long as it exists in the scope and has a name which you can prepend operator&
to. Dereferencing the pointer is a different thing, but that wasn't the question.
Now, the subtle problem is that the standard defines the result of operator&
for non-static struct members as "“pointer to member of class C of type T” and is a prvalue designating C::m".
Which basically means that ptr_str&str
will take the address of str
, but the type is not pointer-to, but pointer-to-member-of. It is then implicitly and silently cast to pointer-to.
In other words, although you do not need to explicitly write &this->str
, that's nevertheless what its type is -- it's what it is and what it means [1].
Is this
valid, and is it safe to use it within the initializer list? Well yes, just... barely. It's safe to use it as long as it's not being used to access uninitialized members or virtual functions, directly or indirectly. Which, as it happens, is the case here (it might not be the case in a different, arguably contrieved case).
[1] Funnily, paragraph 4 starts with a clause that says that no member pointer is formed when you put stuff in parenthese. That's remarkable because most people would probably do that just to be 100% sure they got operator precedence right. But if I read correctly, then
&this->foo
and &(this->foo)
are not in any way the same!Funny question.
It is legitimate and will "work", though barely. There is a little "but" related to types which makes the whole thing a bit awkward with a bad taste (but not illegitimate), and which might make it illegal some border cases involving inheritance.
You can, of course, take the address of any object whether it's initialized or not, as long as it exists in the scope and has a name which you can prepend operator&
to. Dereferencing the pointer is a different thing, but that wasn't the question.
Now, the subtle problem is that the standard defines the result of operator&
for non-static struct members as "“pointer to member of class C of type T” and is a prvalue designating C::m".
Which basically means that ptr_str&str
will take the address of str
, but the type is not pointer-to, but pointer-to-member-of. It is then implicitly and silently cast to pointer-to.
In other words, although you do not need to explicitly write &this->str
, that's nevertheless what its type is -- it's what it is and what it means [1].
Is this
valid, and is it safe to use it within the initializer list? Well yes, just... barely. It's safe to use it as long as it's not being used to access uninitialized members or virtual functions, directly or indirectly. Which, as it happens, is the case here (it might not be the case in a different, arguably contrieved case).
[1] Funnily, paragraph 4 starts with a clause that says that no member pointer is formed when you put stuff in parenthese. That's remarkable because most people would probably do that just to be 100% sure they got operator precedence right. But if I read correctly, then
&this->foo
and &(this->foo)
are not in any way the same!answered May 10 at 17:12
DamonDamon
52.9k15104161
52.9k15104161
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it usesthis
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?
– Ayxan
May 10 at 17:21
1
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not usingthis
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).
– Damon
May 10 at 17:32
2
No,&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write&Klass::str
. Similarly,&this->foo
and&(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.
– Ben Voigt
May 10 at 17:50
2
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
1
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
add a comment |
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it usesthis
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?
– Ayxan
May 10 at 17:21
1
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not usingthis
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).
– Damon
May 10 at 17:32
2
No,&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write&Klass::str
. Similarly,&this->foo
and&(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.
– Ben Voigt
May 10 at 17:50
2
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
1
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it uses
this
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?– Ayxan
May 10 at 17:21
Wow, I had to read it twice. What I understood is that the code I've got there is strictly speaking illegal (since it uses
this
to access an uninitialized member), and it is so subtle that I'd better avoid it altogether. Is that correct?– Ayxan
May 10 at 17:21
1
1
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not using
this
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).– Damon
May 10 at 17:32
No, no, no. It is strictly legal. But it gets very "close" to being illegal. Close, but not over the line. You are not using
this
to access an uninitialized member, you are using it to take the uninitialized member's address. Which is just as much as you are allowed to do legitimately (for complete members of the same class or superclasses, but not subclasses or members of incomplete type). While not wrong, I would personally nevertheless avoid it because it makes me feel funny (legitimate as it may be).– Damon
May 10 at 17:32
2
2
No,
&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write &Klass::str
. Similarly, &this->foo
and &(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.– Ben Voigt
May 10 at 17:50
No,
&
with an unqualified name never creates a pointer-to-member. If you wanted a pointer-to-member, you would have to write &Klass::str
. Similarly, &this->foo
and &(this->foo)
are the same, both create ordinary pointers because these are member-access expressions, not qualified names.– Ben Voigt
May 10 at 17:50
2
2
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
The Standard rule you link to starts out with "If the operand is a qualified-id" Which is why it does not apply here.
– Ben Voigt
May 10 at 17:52
1
1
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
Moreover, there are no implicit conversions from a pointer-to-member type to a pointer type (what object would it bind to?).
– Sneftel
May 10 at 17:54
add a comment |
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Just curious why do you need it?
– vahancho
May 10 at 13:45
1
@vahancho I am adding a private vector to the end of a class (so it is physically separate from the interface) but this vector takes pointers to some public members. I would like to know if I have to move it up or if this is fine.
– Ayxan
May 10 at 13:51
1
just make sure you do not try to access the member before its actually there
– formerlyknownas_463035818
May 10 at 13:56
2
@Ayxan Language-lawyer questions might bring into this site interesting discussions. I would therefore suggest to wait a little longer with such a quick acceptance of the first answer. Though it may be correct, other opinions might be valuable as well.
– Daniel Langr
May 10 at 14:01
@DanielLangr nice suggestion.
– Ayxan
May 10 at 14:03