[SOLVED] C++ : function adress/ Variable adress

LoDam · 03-29-2011, 05:25 AM

Hi LQ,

I have a question regarding function adress.
Maybe it is easier to understand my problem with a bit of code:

Code:

#include <iostream>
  2 using namespace std;
  3 void func(void)
  4 {
  5     cout << "Did you call me?" << endl;
  6 return;
  7 }
  8 int main(int argc,char* argv[])
  9 {
 10     void (*f)() = &func;
 11     int i = 0;
 12 
 13     cout
 14         << "without casting :" << endl
 15         << "f: " << f  << " main: " << main << endl
 16         << "with casting :" << endl
 17         << "f:  " << (long)f << " main: " << (long)main
 18         << endl;
 19     cout << "With variables: " << endl << &i << " " << long(&i) << endl;
 20 return 0;
 21 }

and the output I get :

Code:

[Alpha][Tests] ./test_pointfunc
without casting :
f: 1 main: 1
with casting :
f:  4196516 main: 4196550
With variables: 
0x7fff10662df4 140733468519924

Why do I get a simple '1' for the function adress when not casted to couble? (while I get an hex adress for a variable). Also, why is it the same result for all the function (I did it here with the main, but if I had another function, I get the same results).

Is it machine specific, is it that I did something wrong, or is it something I didn't understand?

Cheers,
Loic.

dwhitney67 · 03-29-2011, 05:41 AM

The function pointer is interpreted as a bool when passed to the ostream object. You can verify this with:

Code:

#include <iostream>

void func(void)
{
}

int main()
{
   void (*f)() = func;
   void (*g)() = NULL;

   std::cout << "f = " << f << std::endl;
   std::cout << "g = " << g << std::endl;
}

Suffice to say, your question has been asked before; read here for the response.