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Recursive calls to a function - why is the address of the parameter passed to it lowering with each call?
What is the direction of stack growth in most modern systems?Why isn't sizeof for a struct equal to the sum of sizeof of each member?How to call a parent class function from derived class function?Why do we need virtual functions in C++?Pretty-print C++ STL containersAre the days of passing const std::string & as a parameter over?Recursive Reverse FunctionWhy can I not move unique_ptr from a set to a function argument using an iterator?Why can I not call reserve on a vector of const elements?recursive variadic template function call “loses” pointer on second argument typeHaving issues with .h file, it doesn't seem to be linking correctly
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
10
Consider following code:
#include <iostream>
using namespace std;
void test_func(int address)
cout<<&address<<" ";
if(address < 0x7FFBEE26)
test_func(address);
int main()
test_func(512);
cout<<"Hello";
return 0;
Hello from main() is certainly not reached, since the recursive calls to test_func never end.
However, from what I can see in the cout present in test_func - the addresses being printed are lower and lower with each iteration. Why is that happening?
c++ recursion memory-address
edited 2 days ago
Muntasir
6381919
asked Apr 22 at 20:48
tears allotears allo
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 1 more comment
10
Consider following code:
#include <iostream>
using namespace std;
void test_func(int address)
cout<<&address<<" ";
if(address < 0x7FFBEE26)
test_func(address);
int main()
test_func(512);
cout<<"Hello";
return 0;
Hello from main() is certainly not reached, since the recursive calls to test_func never end.
However, from what I can see in the cout present in test_func - the addresses being printed are lower and lower with each iteration. Why is that happening?
c++ recursion memory-address
edited 2 days ago
Muntasir
6381919
asked Apr 22 at 20:48
tears allotears allo
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
You are passing a copy - that has to have an address
– UnholySheep
Apr 22 at 20:50
1
Remember that the default stack size on linux is 10MB and its 1 MB on windows. Also the stack need not be in the same location each time you run your program.
– drescherjm
Apr 22 at 21:04
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_funcis the last line in the function...
– cyberbisson
Apr 22 at 21:33
9
@cyberbisson The parameters of the nested invocations oftest_funcmust appear to have different addresses per language rules, and because the address ofaddresswas passed tooperator<<the compiler can't prove that this is unobservable.
– T.C.
Apr 22 at 22:46
1
@Deduplicator Yes.
– T.C.
Apr 22 at 23:48
|
show 1 more comment
10
10
10
1
Consider following code:
#include <iostream>
using namespace std;
void test_func(int address)
cout<<&address<<" ";
if(address < 0x7FFBEE26)
test_func(address);
int main()
test_func(512);
cout<<"Hello";
return 0;
Hello from main() is certainly not reached, since the recursive calls to test_func never end.
However, from what I can see in the cout present in test_func - the addresses being printed are lower and lower with each iteration. Why is that happening?
c++ recursion memory-address
edited 2 days ago
Muntasir
6381919
asked Apr 22 at 20:48
tears allotears allo
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider following code:
#include <iostream>
using namespace std;
void test_func(int address)
cout<<&address<<" ";
if(address < 0x7FFBEE26)
test_func(address);
int main()
test_func(512);
cout<<"Hello";
return 0;
Hello from main() is certainly not reached, since the recursive calls to test_func never end.
However, from what I can see in the cout present in test_func - the addresses being printed are lower and lower with each iteration. Why is that happening?
c++ recursion memory-address
c++ recursion memory-address
edited 2 days ago
Muntasir
6381919
asked Apr 22 at 20:48
tears allotears allo
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Muntasir
6381919
asked Apr 22 at 20:48
tears allotears allo
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Muntasir
6381919
edited 2 days ago
Muntasir
6381919
edited 2 days ago
Muntasir
6381919
6381919
asked Apr 22 at 20:48
tears allotears allo
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 22 at 20:48
tears allotears allo
594
asked Apr 22 at 20:48
tears allotears allo
594
594
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tears allo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
You are passing a copy - that has to have an address
– UnholySheep
Apr 22 at 20:50
1
Remember that the default stack size on linux is 10MB and its 1 MB on windows. Also the stack need not be in the same location each time you run your program.
– drescherjm
Apr 22 at 21:04
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_funcis the last line in the function...
– cyberbisson
Apr 22 at 21:33
9
@cyberbisson The parameters of the nested invocations oftest_funcmust appear to have different addresses per language rules, and because the address ofaddresswas passed tooperator<<the compiler can't prove that this is unobservable.
– T.C.
Apr 22 at 22:46
1
@Deduplicator Yes.
– T.C.
Apr 22 at 23:48
|
show 1 more comment
4
You are passing a copy - that has to have an address
– UnholySheep
Apr 22 at 20:50
1
Remember that the default stack size on linux is 10MB and its 1 MB on windows. Also the stack need not be in the same location each time you run your program.
– drescherjm
Apr 22 at 21:04
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_funcis the last line in the function...
– cyberbisson
Apr 22 at 21:33
9
@cyberbisson The parameters of the nested invocations oftest_funcmust appear to have different addresses per language rules, and because the address ofaddresswas passed tooperator<<the compiler can't prove that this is unobservable.
– T.C.
Apr 22 at 22:46
1
@Deduplicator Yes.
– T.C.
Apr 22 at 23:48
4
4
You are passing a copy - that has to have an address
– UnholySheep
Apr 22 at 20:50
You are passing a copy - that has to have an address
– UnholySheep
Apr 22 at 20:50
1
1
Remember that the default stack size on linux is 10MB and its 1 MB on windows. Also the stack need not be in the same location each time you run your program.
– drescherjm
Apr 22 at 21:04
Remember that the default stack size on linux is 10MB and its 1 MB on windows. Also the stack need not be in the same location each time you run your program.
– drescherjm
Apr 22 at 21:04
I can't understand why this isn't eligible for tail-call optimization. The invocation of
test_func is the last line in the function...– cyberbisson
Apr 22 at 21:33
I can't understand why this isn't eligible for tail-call optimization. The invocation of
test_func is the last line in the function...– cyberbisson
Apr 22 at 21:33
9
9
@cyberbisson The parameters of the nested invocations of
test_func must appear to have different addresses per language rules, and because the address of address was passed to operator<< the compiler can't prove that this is unobservable.– T.C.
Apr 22 at 22:46
@cyberbisson The parameters of the nested invocations of
test_func must appear to have different addresses per language rules, and because the address of address was passed to operator<< the compiler can't prove that this is unobservable.– T.C.
Apr 22 at 22:46
1
1
@Deduplicator Yes.
– T.C.
Apr 22 at 23:48
@Deduplicator Yes.
– T.C.
Apr 22 at 23:48
|
show 1 more comment
1 Answer
1
active
oldest
votes
22
Likely address is being placed on the stack and, on your platform, the stack grows downward in memory. See this question about stack growth direction for more.
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
3
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers iscdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results
– Remy Lebeau
Apr 23 at 2:55
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
It's difficult to putaddressin a register because there will be many instances ofaddressat the same time, one for each instance oftest_functhat's in the process of executing.
– David Schwartz
2 days ago
add a comment |
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22
Likely address is being placed on the stack and, on your platform, the stack grows downward in memory. See this question about stack growth direction for more.
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
3
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers iscdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results
– Remy Lebeau
Apr 23 at 2:55
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
It's difficult to putaddressin a register because there will be many instances ofaddressat the same time, one for each instance oftest_functhat's in the process of executing.
– David Schwartz
2 days ago
add a comment |
22
Likely address is being placed on the stack and, on your platform, the stack grows downward in memory. See this question about stack growth direction for more.
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
3
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers iscdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results
– Remy Lebeau
Apr 23 at 2:55
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
It's difficult to putaddressin a register because there will be many instances ofaddressat the same time, one for each instance oftest_functhat's in the process of executing.
– David Schwartz
2 days ago
add a comment |
22
22
22
Likely address is being placed on the stack and, on your platform, the stack grows downward in memory. See this question about stack growth direction for more.
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
Likely address is being placed on the stack and, on your platform, the stack grows downward in memory. See this question about stack growth direction for more.
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
answered Apr 22 at 20:50
David SchwartzDavid Schwartz
140k14147232
140k14147232
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
3
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers iscdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results
– Remy Lebeau
Apr 23 at 2:55
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
It's difficult to putaddressin a register because there will be many instances ofaddressat the same time, one for each instance oftest_functhat's in the process of executing.
– David Schwartz
2 days ago
add a comment |
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
3
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers iscdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results
– Remy Lebeau
Apr 23 at 2:55
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
It's difficult to putaddressin a register because there will be many instances ofaddressat the same time, one for each instance oftest_functhat's in the process of executing.
– David Schwartz
2 days ago
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
Apr 23 at 2:35
3
3
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers is
cdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results– Remy Lebeau
Apr 23 at 2:55
@ᆼᆺᆼ no, it is because on 32bit systems, the default calling convention in most C/C++ compilers is
cdecl, which passes parameters on the call stack only. Compile your code for 64bit, or alter your function to use a register-based calling convention, and you will likely see different results– Remy Lebeau
Apr 23 at 2:55
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@RemyLebeau What even is the address of a passed-by-register argument? Apparently, GCC artificially moves the register value onto the stack and then takes the address: godbolt.org/z/LRz5DS
– ComFreek
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
@ComFreek registers don't have addresses. Copying a register value to a memory block is the only way to get a memory address.
– Remy Lebeau
2 days ago
It's difficult to put
address in a register because there will be many instances of address at the same time, one for each instance oftest_func that's in the process of executing.– David Schwartz
2 days ago
It's difficult to put
address in a register because there will be many instances of address at the same time, one for each instance oftest_func that's in the process of executing.– David Schwartz
2 days ago
add a comment |
tears allo is a new contributor. Be nice, and check out our Code of Conduct.
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tears allo is a new contributor. Be nice, and check out our Code of Conduct.
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4
You are passing a copy - that has to have an address
– UnholySheep
Apr 22 at 20:50
1
Remember that the default stack size on linux is 10MB and its 1 MB on windows. Also the stack need not be in the same location each time you run your program.
– drescherjm
Apr 22 at 21:04
I can't understand why this isn't eligible for tail-call optimization. The invocation of
test_funcis the last line in the function...– cyberbisson
Apr 22 at 21:33
9
@cyberbisson The parameters of the nested invocations of
test_funcmust appear to have different addresses per language rules, and because the address ofaddresswas passed tooperator<<the compiler can't prove that this is unobservable.– T.C.
Apr 22 at 22:46
1
@Deduplicator Yes.
– T.C.
Apr 22 at 23:48