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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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 allotears 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.
|
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?
c++
edited 6 hours ago
drescherjm
6,59923553
asked 7 hours ago
tears allotears 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
7 hours ago
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_funcis the last line in the function...
– cyberbisson
6 hours ago
6
@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.
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?
c++
edited 6 hours ago
drescherjm
6,59923553
asked 7 hours ago
tears allotears 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.
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++
edited 6 hours ago
drescherjm
6,59923553
asked 7 hours ago
tears allotears 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.
edited 6 hours ago
drescherjm
6,59923553
asked 7 hours ago
tears allotears 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.
edited 6 hours ago
drescherjm
6,59923553
edited 6 hours ago
drescherjm
6,59923553
edited 6 hours ago
drescherjm
6,59923553
6,59923553
asked 7 hours ago
tears allotears 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.
asked 7 hours ago
tears allotears allo
491
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 oftest_funcis the last line in the function...
– cyberbisson
6 hours ago
6
@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.
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
7 hours ago
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_funcis the last line in the function...
– cyberbisson
6 hours ago
6
@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.
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
|
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 SchwartzDavid 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 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
1 hour ago
add a comment |
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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 SchwartzDavid 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 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
1 hour 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 SchwartzDavid 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 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
1 hour 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 SchwartzDavid 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 SchwartzDavid Schwartz
140k14146232
answered 7 hours ago
David SchwartzDavid Schwartz
140k14146232
answered 7 hours ago
David SchwartzDavid Schwartz
140k14146232
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 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
1 hour 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 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
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
add a comment |
tears allo is a new contributor. Be nice, and check out our Code of Conduct.
tears allo is a new contributor. Be nice, and check out our Code of Conduct.
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
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_funcis the last line in the function...– cyberbisson
6 hours ago
6
@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.
5 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
4 hours ago