How can I print value of a function pointer in C++?

Asked Jun 14 '18 at 15:32

Active Jun 14 '18 at 15:32

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I have a question here and do not understand why I can't print the memory address of my function pointer. Can anyone explain why? Thanks!

#include <stdio.h>
#include <iostream>

void displayFourIncrementVar(int &x)
{
    printf( "%d\n", 4 );
    x++;
}

int main()
{
    int y = 4;
    int *z = &y;
    std::cout << *z << std::endl;
    std::cout << z << std::endl << std::endl;
    //printing out z we have an address.  

    void (*garbage)(int&) = &displayFourIncrementVar;
    std::cout << garbage << std::endl << std::endl;
    //Printing out garbage we have no address, instead we get 1, why's that?
    //I thought pointers store memory addresses in dexadecimal.

    int x = 2;
    (*garbage)(x);
    std::cout << x << std::endl; 
    return 0;
}

asked Jun 14 '18 at 15:32

Fortunata

what is it printing? – Luke Xu Jun 14 '18 at 15:32
It prints 1 ... – Fortunata Jun 14 '18 at 15:33
Am I not understanding something here? I believe garbage should hold a memory address but instead it prints out 1. – Fortunata Jun 14 '18 at 15:36
A function pointer is not convertible to `void*`, which is the overload used by `std::ostream` to print pointers. The only matching overload seems to be `bool`. Though your code seems to work fine for me on Visual Studio 2015, I don't think it's supposed to. – François Andrieux Jun 14 '18 at 15:37
use printf. printf"%p", garbage) – Atul Jun 14 '18 at 16:31

How can I print value of a function pointer in C++?

0 Answers0