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

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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?

edited 6 hours ago

drescherjm

6,59923553

asked 7 hours ago

tears allo

491

New contributor

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
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

@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

|
show 2 more comments

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?

edited 6 hours ago

drescherjm

6,59923553

asked 7 hours ago

tears allo

491

New contributor

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
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

@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

|
show 2 more comments

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?

edited 6 hours ago

drescherjm

6,59923553

asked 7 hours ago

tears allo

491

New contributor

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++

edited 6 hours ago

drescherjm

6,59923553

asked 7 hours ago

tears allo

491

New contributor

edited 6 hours ago

drescherjm

6,59923553

asked 7 hours ago

tears allo

491

New contributor

edited 6 hours ago

drescherjm

6,59923553

edited 6 hours ago

drescherjm

6,59923553

edited 6 hours ago

drescherjm

6,59923553

asked 7 hours ago

tears allo

491

New contributor

asked 7 hours ago

tears allo

491

asked 7 hours ago

tears allo

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.

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
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

@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

|
show 2 more comments

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
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

@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

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

– UnholySheep
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
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

@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

|
show 2 more comments

1 Answer
1

active

oldest

votes

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 7 hours ago

David Schwartz

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
55 mins ago

add a comment |

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1 Answer
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1 Answer
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oldest

votes

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 7 hours ago

David Schwartz

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
55 mins ago

add a comment |

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 7 hours ago

David Schwartz

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
55 mins ago

add a comment |

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 7 hours ago

David Schwartz

140k14146232

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 7 hours ago

David Schwartz

140k14146232

answered 7 hours ago

David Schwartz

140k14146232

answered 7 hours ago

David Schwartz

140k14146232

answered 7 hours ago

David Schwartz

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
55 mins ago

add a comment |

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
55 mins 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
55 mins ago

add a comment |

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