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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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++
New contributor
|
show 1 more comment
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++
New contributor
4
You are passing a copy - that has to have an address
– UnholySheep
8 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
8 hours ago
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_func
is the last line in the function...
– cyberbisson
7 hours ago
6
@cyberbisson The parameters of the nested invocations oftest_func
must appear to have different addresses per language rules, and because the address ofaddress
was passed tooperator<<
the compiler can't prove that this is unobservable.
– T.C.
6 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
6 hours ago
|
show 1 more comment
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++
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++
c++
New contributor
New contributor
edited 8 hours ago
drescherjm
6,59923553
6,59923553
New contributor
asked 8 hours ago
tears allotears allo
471
471
New contributor
New contributor
4
You are passing a copy - that has to have an address
– UnholySheep
8 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
8 hours ago
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_func
is the last line in the function...
– cyberbisson
7 hours ago
6
@cyberbisson The parameters of the nested invocations oftest_func
must appear to have different addresses per language rules, and because the address ofaddress
was passed tooperator<<
the compiler can't prove that this is unobservable.
– T.C.
6 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
6 hours ago
|
show 1 more comment
4
You are passing a copy - that has to have an address
– UnholySheep
8 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
8 hours ago
I can't understand why this isn't eligible for tail-call optimization. The invocation oftest_func
is the last line in the function...
– cyberbisson
7 hours ago
6
@cyberbisson The parameters of the nested invocations oftest_func
must appear to have different addresses per language rules, and because the address ofaddress
was passed tooperator<<
the compiler can't prove that this is unobservable.
– T.C.
6 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
6 hours ago
4
4
You are passing a copy - that has to have an address
– UnholySheep
8 hours ago
You are passing a copy - that has to have an address
– UnholySheep
8 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
8 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
8 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
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
7 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.
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.
6 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
6 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
6 hours ago
|
show 1 more comment
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.
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
1
@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
2 hours ago
add a comment |
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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.
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
1
@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
2 hours 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.
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
1
@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
2 hours 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.
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 8 hours ago
David SchwartzDavid Schwartz
140k14146232
140k14146232
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
1
@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
2 hours ago
add a comment |
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
1
@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
2 hours ago
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
Is it placed on the stack instead of in a register because its address is taken?
– ᆼᆺᆼ
2 hours ago
1
1
@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
2 hours 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
2 hours 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.
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
8 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
8 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
7 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 ofaddress
was passed tooperator<<
the compiler can't prove that this is unobservable.– T.C.
6 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
6 hours ago