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Does it make sense for a function to return an rvalue reference?


Is returning by rvalue reference more efficient?What should main() return in C and C++?Advantages of using forwardClasses, Rvalues and Rvalue ReferencesImage Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhy should I use a pointer rather than the object itself?Does this rvalue signature pattern make sense?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsReturn value or rvalue reference?Rvalue and Lvalue Referencesrvalue reference or forwarding reference?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








18















What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?









share|improve this question



















  • 6





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    May 2 at 19:32







  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    May 2 at 19:34






  • 1





    See this answer.

    – lubgr
    May 2 at 19:37











  • Isn't that a forwarding reference?

    – user2357112
    May 3 at 1:52

















18















What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?









share|improve this question



















  • 6





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    May 2 at 19:32







  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    May 2 at 19:34






  • 1





    See this answer.

    – lubgr
    May 2 at 19:37











  • Isn't that a forwarding reference?

    – user2357112
    May 3 at 1:52













18












18








18


3






What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?









share|improve this question
















What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?






c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 3 at 2:35









Boann

37.7k1291123




37.7k1291123










asked May 2 at 19:22









Martin B.Martin B.

842314




842314







  • 6





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    May 2 at 19:32







  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    May 2 at 19:34






  • 1





    See this answer.

    – lubgr
    May 2 at 19:37











  • Isn't that a forwarding reference?

    – user2357112
    May 3 at 1:52












  • 6





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    May 2 at 19:32







  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    May 2 at 19:34






  • 1





    See this answer.

    – lubgr
    May 2 at 19:37











  • Isn't that a forwarding reference?

    – user2357112
    May 3 at 1:52







6




6





std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

– François Andrieux
May 2 at 19:32






std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

– François Andrieux
May 2 at 19:32





1




1





@FrançoisAndrieux the std::get family of functions, too.

– Brian
May 2 at 19:34





@FrançoisAndrieux the std::get family of functions, too.

– Brian
May 2 at 19:34




1




1





See this answer.

– lubgr
May 2 at 19:37





See this answer.

– lubgr
May 2 at 19:37













Isn't that a forwarding reference?

– user2357112
May 3 at 1:52





Isn't that a forwarding reference?

– user2357112
May 3 at 1:52












3 Answers
3






active

oldest

votes


















12














For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() return bar;
;


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() & return bar;
std::vector<int>&& get_vec() && return std::move(bar);
;


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer























  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    May 2 at 19:37











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    May 2 at 19:38











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    May 2 at 19:39






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    May 2 at 20:03







  • 2





    std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    May 2 at 20:08



















3















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.




But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper 

auto operator*() & -> Heavy&
return heavy;


auto operator*() && -> Heavy&&
return std::move(heavy);


private:
Heavy instance;
;

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer

























  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:51











  • @CruzJean corrected

    – Guillaume Racicot
    May 2 at 20:57


















0














I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger

public:
void log(const char* msg)
logs.append(msg);

std::vector<std::string>&& dumpLogs()
return std::move(logs);

private:
std::vector<std::string> logs;
;


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs()
auto dumped_logs = logs;
return dumped_logs;






share|improve this answer























  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:52












  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    May 2 at 20:58











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    May 2 at 22:25











  • @CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

    – Quimby
    2 days ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









12














For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() return bar;
;


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() & return bar;
std::vector<int>&& get_vec() && return std::move(bar);
;


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer























  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    May 2 at 19:37











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    May 2 at 19:38











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    May 2 at 19:39






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    May 2 at 20:03







  • 2





    std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    May 2 at 20:08
















12














For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() return bar;
;


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() & return bar;
std::vector<int>&& get_vec() && return std::move(bar);
;


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer























  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    May 2 at 19:37











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    May 2 at 19:38











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    May 2 at 19:39






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    May 2 at 20:03







  • 2





    std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    May 2 at 20:08














12












12








12







For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() return bar;
;


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() & return bar;
std::vector<int>&& get_vec() && return std::move(bar);
;


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer













For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() return bar;
;


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo

std::vector<int> bar
public:
foo(int n) : bar(n)
std::vector<int>& get_vec() & return bar;
std::vector<int>&& get_vec() && return std::move(bar);
;


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.







share|improve this answer












share|improve this answer



share|improve this answer










answered May 2 at 19:34









NathanOliverNathanOliver

101k17141224




101k17141224












  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    May 2 at 19:37











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    May 2 at 19:38











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    May 2 at 19:39






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    May 2 at 20:03







  • 2





    std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    May 2 at 20:08


















  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    May 2 at 19:37











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    May 2 at 19:38











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    May 2 at 19:39






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    May 2 at 20:03







  • 2





    std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    May 2 at 20:08

















@FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

– NathanOliver
May 2 at 19:37





@FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

– NathanOliver
May 2 at 19:37













Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

– super
May 2 at 19:38





Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

– super
May 2 at 19:38













@super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

– NathanOliver
May 2 at 19:39





@super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

– NathanOliver
May 2 at 19:39




1




1





@NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

– Cruz Jean
May 2 at 20:03






@NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

– Cruz Jean
May 2 at 20:03





2




2





std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

– Yakk - Adam Nevraumont
May 2 at 20:08






std::vector<int> get_vec() && return std::move(bar); in this case ends up being better 999/1000. Can you come up with a better example?

– Yakk - Adam Nevraumont
May 2 at 20:08














3















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.




But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper 

auto operator*() & -> Heavy&
return heavy;


auto operator*() && -> Heavy&&
return std::move(heavy);


private:
Heavy instance;
;

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer

























  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:51











  • @CruzJean corrected

    – Guillaume Racicot
    May 2 at 20:57















3















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.




But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper 

auto operator*() & -> Heavy&
return heavy;


auto operator*() && -> Heavy&&
return std::move(heavy);


private:
Heavy instance;
;

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer

























  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:51











  • @CruzJean corrected

    – Guillaume Racicot
    May 2 at 20:57













3












3








3








T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.




But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper 

auto operator*() & -> Heavy&
return heavy;


auto operator*() && -> Heavy&&
return std::move(heavy);


private:
Heavy instance;
;

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer
















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.




But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper 

auto operator*() & -> Heavy&
return heavy;


auto operator*() && -> Heavy&&
return std::move(heavy);


private:
Heavy instance;
;

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());






share|improve this answer














share|improve this answer



share|improve this answer








edited May 2 at 20:57

























answered May 2 at 19:35









Guillaume RacicotGuillaume Racicot

16.9k53874




16.9k53874












  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:51











  • @CruzJean corrected

    – Guillaume Racicot
    May 2 at 20:57

















  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:51











  • @CruzJean corrected

    – Guillaume Racicot
    May 2 at 20:57
















In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
May 2 at 20:51





In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
May 2 at 20:51













@CruzJean corrected

– Guillaume Racicot
May 2 at 20:57





@CruzJean corrected

– Guillaume Racicot
May 2 at 20:57











0














I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger

public:
void log(const char* msg)
logs.append(msg);

std::vector<std::string>&& dumpLogs()
return std::move(logs);

private:
std::vector<std::string> logs;
;


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs()
auto dumped_logs = logs;
return dumped_logs;






share|improve this answer























  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:52












  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    May 2 at 20:58











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    May 2 at 22:25











  • @CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

    – Quimby
    2 days ago















0














I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger

public:
void log(const char* msg)
logs.append(msg);

std::vector<std::string>&& dumpLogs()
return std::move(logs);

private:
std::vector<std::string> logs;
;


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs()
auto dumped_logs = logs;
return dumped_logs;






share|improve this answer























  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:52












  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    May 2 at 20:58











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    May 2 at 22:25











  • @CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

    – Quimby
    2 days ago













0












0








0







I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger

public:
void log(const char* msg)
logs.append(msg);

std::vector<std::string>&& dumpLogs()
return std::move(logs);

private:
std::vector<std::string> logs;
;


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs()
auto dumped_logs = logs;
return dumped_logs;






share|improve this answer













I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger

public:
void log(const char* msg)
logs.append(msg);

std::vector<std::string>&& dumpLogs()
return std::move(logs);

private:
std::vector<std::string> logs;
;


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs()
auto dumped_logs = logs;
return dumped_logs;







share|improve this answer












share|improve this answer



share|improve this answer










answered May 2 at 19:32









QuimbyQuimby

1,297714




1,297714












  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:52












  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    May 2 at 20:58











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    May 2 at 22:25











  • @CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

    – Quimby
    2 days ago

















  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    May 2 at 20:52












  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    May 2 at 20:58











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    May 2 at 22:25











  • @CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

    – Quimby
    2 days ago
















That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
May 2 at 20:52






That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
May 2 at 20:52














@CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

– Quimby
May 2 at 20:58





@CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

– Quimby
May 2 at 20:58













return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

– Cruz Jean
May 2 at 22:25





return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

– Cruz Jean
May 2 at 22:25













@CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

– Quimby
2 days ago





@CruzJean Do you mean e.g. auto&& x = Logger.dumpLogs(); ? Yes, that is dangling reference, but that happens for all getters no matter what type of reference they return. My use case would be something like Logger l; /*calls to log()*/, auto logs = l.dumpLogs();/*log again.. and repeat.*/

– Quimby
2 days ago

















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