Address of function pointers in C++ [duplicate] - c++

Question

This question already has answers here:
How to print function pointers with cout?
(7 answers)
Closed 9 years ago.
I'm not clear on what the values that are being returning from calling:
&next, fp, *fp, &return_func_ptr, fp_ptr, &fp_ptr, *fp_ptr
They all seem to give me the value 1. What does it mean?
Also, how would I declare
int (*return_f())(char)
to receive a parameter without using typedef?
#include <iostream>
int next(int n){
return n+99;
}
// returns pointer to a function
typedef int (*fptr)(int); // using typdef
fptr return_func_ptr(){
return next;
}
int f(char){
return 0;
}
int (*return_f())(char){ // how do you pass a parameter here?
// std::cout << "do something with " << param << std::endl;
return f;
}
int main()
{
int x = 5;
// p points to x
int *p = &x;
std::cout << "x=" << x << std::endl; // 5, value of x
std::cout << "&x=" << &x << std::endl; // 0x7fff6447a82c, address of x
std::cout << "p=" << p << std::endl; // 0x7fff6447a82c, value of p is address of x
std::cout << "*p=" << *p << std::endl; // 5, value of x (p dereferenced)
std::cout << "&p=" << &p << std::endl; // 0x7fff6447a820, address of p pointer
// change value of x thru p
// p = 6; // error, can't set int* to int
*p = 6;
std::cout << "x=" << x << std::endl; // 6
int y = 2;
// int *q = y; // error can't initiate with type int, needs int*
// pointer to a function
int (*fp)(int);
std::cout << "&fp=" << &fp << std::endl; // 0x7fff66da6810, address of pointer fp
std::cout << "fp=" << fp << std::endl; // 0, value of pointer fp
fp = &next; // fp points to function next(int)
fp = next;
std::cout << "&next=" << &next << std::endl; // 1, address of function?
std::cout << "fp=" << fp << std::endl; // 1, value is address of function?
std::cout << "&fp=" << &fp << std::endl; // 0x7fff66da6810, address of pointer fp?
std::cout << "*fp=" << *fp << std::endl; // 1, address of function?
// calling function thru pointer
int i = 0;
i = (*fp)(i);
std::cout << "i=" << i << std::endl; // 99
i = fp(i);
std::cout << "i=" << i << std::endl; // 198
// function returns pointer to function
fptr fp_ptr = return_func_ptr();
std::cout << "&return_func_ptr=" << &return_func_ptr << std::endl; // 1
std::cout << "fp_ptr=" << *fp_ptr << std::endl; // 1
std::cout << "&fp_ptr=" << *fp_ptr << std::endl; // 1
std::cout << "*fp_ptr=" << *fp_ptr << std::endl; // 1
int j = fp_ptr(1);
std::cout << "j=" << j << std::endl; // 100
}

Answer 1

There is some pointer here who seems not clear :
// pointer to a function
int (*fp)(int);
std::cout << "&fp=" << &fp << std::endl; // 0x7fff66da6810, address of pointer fp
std::cout << "fp=" << fp << std::endl; // 0, value of pointer fp
Here fp is undefined. Those lines have an undefined behaviour.
After that :
// function returns pointer to function
fptr fp_ptr = return_func_ptr();
std::cout << "&return_func_ptr=" << &return_func_ptr << std::endl; // 1
std::cout << "fp_ptr=" << *fp_ptr << std::endl; // 1
std::cout << "&fp_ptr=" << *fp_ptr << std::endl; // 1
// ^^^^^^^^^^ ^^^^^^^
std::cout << "*fp_ptr=" << *fp_ptr << std::endl; // 1
There are two things here :
On the line I pointed, I'm not sure it is what you wanted to test.
Also, cout doesn't have an overload to take a function pointer, it will take a bool instead. So it should be :
std::cout << "fn_ptr=" << reinterpret_cast<void*>( fn_ptr ) << std::endl;
I would suggest you to read this article about function pointer, it explains almost all you need to know : http://www.learncpp.com/cpp-tutorial/78-function-pointers/

Answer 2

std::cout << "fp_ptr=" << *fp_ptr << std::endl;
should be
std::cout << "fp_ptr=" << (void*)fp_ptr << std::endl;
The cout operator doesn't have an overload for a function pointer, so it uses bool instead. That's why you always get 1 as output. When I compile your code, I even get a warning for that, telling me that it will always evaluate to true. You should switch on all warnings and try to get rid of them.