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Recursive calls to a function - why is the address of the parameter passed to it lowering with each call?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!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 containersHow to pass normal param as well as template param in a template function in C++?Are 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?Having issues with .h file, it doesn't seem to be linking correctly
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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++
edited 7 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
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++
edited 7 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
5 hours ago
|
show 2 more comments
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++
edited 7 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 7 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 7 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 7 hours ago
drescherjm
6,59923553
edited 7 hours ago
drescherjm
6,59923553
edited 7 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
5 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
5 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
5 hours ago
@T.C. So, the problem is that the callee might remember and use it still?
– Deduplicator
5 hours ago
|
show 2 more comments
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.
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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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.
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 |
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.
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 |
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.
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 |
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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
5 hours ago