Recursive calls to a function - why is the address of the parameter passed to it lowering with each call? ...

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Recursive calls to a function - why is the address of the parameter passed to it lowering with each call?



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9















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++







share|improve this question









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
    7 hours ago






  • 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
    7 hours ago













  • 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
    6 hours ago








  • 6





    @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.
    5 hours ago











  • @T.C. So, the problem is that the callee might remember and use it still?

    – Deduplicator
    4 hours ago


















9















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++







share|improve this question









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
    7 hours ago






  • 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
    7 hours ago













  • 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
    6 hours ago








  • 6





    @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.
    5 hours ago











  • @T.C. So, the problem is that the callee might remember and use it still?

    – Deduplicator
    4 hours ago














9












9








9








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++







share|improve this question









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++




c++






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 6 hours ago









drescherjm

6,59923553




6,59923553






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 7 hours ago









tears allotears allo

491




491




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
    7 hours ago






  • 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
    7 hours ago













  • 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
    6 hours ago








  • 6





    @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.
    5 hours ago











  • @T.C. So, the problem is that the callee might remember and use it still?

    – Deduplicator
    4 hours ago














  • 4





    You are passing a copy - that has to have an address

    – UnholySheep
    7 hours ago






  • 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
    7 hours ago













  • 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
    6 hours ago








  • 6





    @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.
    5 hours ago











  • @T.C. So, the problem is that the callee might remember and use it still?

    – Deduplicator
    4 hours ago








4




4





You are passing a copy - that has to have an address

– UnholySheep
7 hours ago





You are passing a copy - that has to have an address

– UnholySheep
7 hours ago




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
7 hours ago







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
7 hours ago















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
6 hours ago







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
6 hours ago






6




6





@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.
5 hours ago





@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.
5 hours ago













@T.C. So, the problem is that the callee might remember and use it still?

– Deduplicator
4 hours ago





@T.C. So, the problem is that the callee might remember and use it still?

– Deduplicator
4 hours ago












1 Answer
1






active

oldest

votes


















18














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.






share|improve this answer
























  • Is it placed on the stack instead of in a register because its address is taken?

    – ᆼᆺᆼ
    1 hour ago











  • @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
    1 hour ago














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active

oldest

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active

oldest

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active

oldest

votes









18














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.






share|improve this answer
























  • Is it placed on the stack instead of in a register because its address is taken?

    – ᆼᆺᆼ
    1 hour ago











  • @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
    1 hour ago


















18














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.






share|improve this answer
























  • Is it placed on the stack instead of in a register because its address is taken?

    – ᆼᆺᆼ
    1 hour ago











  • @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
    1 hour ago
















18












18








18







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.






share|improve this answer













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.







share|improve this answer












share|improve this answer



share|improve this answer










answered 7 hours ago









David SchwartzDavid Schwartz

140k14146232




140k14146232













  • Is it placed on the stack instead of in a register because its address is taken?

    – ᆼᆺᆼ
    1 hour ago











  • @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
    1 hour ago





















  • Is it placed on the stack instead of in a register because its address is taken?

    – ᆼᆺᆼ
    1 hour ago











  • @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
    1 hour ago



















Is it placed on the stack instead of in a register because its address is taken?

– ᆼᆺᆼ
1 hour ago





Is it placed on the stack instead of in a register because its address is taken?

– ᆼᆺᆼ
1 hour ago













@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
1 hour ago







@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
1 hour ago














tears allo is a new contributor. Be nice, and check out our Code of Conduct.










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