c++ pointer to function not changed

Question

I have defined some functions and I print their address like this:

#include<iostream>
#include <string>

using std::cout;

std::string func()
{
    return "hello world\n";

}

int func2(int n)
{
    if (n==0)
    {
        cout << func2 << std::endl;
        return 1;
    }

    cout << func2 << std::endl;

    return n + func2(n - 1);
}

//================================================
int main()
{
    int (*fun)(int) = func2;

    cout << fun;

    cout << std::endl << func2(3);
}

When I print the function's name (address) they all print 1 on my compiler (Mingw gcc 4.8 ).

Is it OK or it should differ?

Dimitrios Bouzas · Accepted Answer · 2015-12-13 22:06:21Z

3

It doesn't exist an overload of operator<< for std::ostream that takes a function pointer. Thus the operator<<(std::ostream&, bool) overload is prefered. The address of a function is always evaluated to true when converted to bool. Thus, 1 is printed.

Alternatevely, if a function pointer is not larger than the size of a data pointer, you could cast your function pointer to a void* via reinterpret_cast and evoke the operator<<(std::ostream&, void*) overload and thus get the actual address of the function printed.

int (*fun)(int) = func2;
std::cout << reinterpret_cast<void*>(fun) << std::endl;

Live Demo

However, as correctly Neil and M.M mentioned in the comments there's no standard conversion from a function pointer to a data pointer, and this could evoke undefined behaviour.

Alternatively, and in my humble opinion properly, you could format your function pointer as a char array buffer and convert its address to a string in the following way:

unsigned char *p = reinterpret_cast<unsigned char*>(&func2);
std::stringstream ss;
ss << std::hex << std::setfill('0');
for(int i(sizeof(func2) - 1); i >= 0; --i) ss << std::setw(2) 
                                              << static_cast<unsigned int>(p[i]);
std::cout << ss.str() << std::endl;

Live Demo

edited Dec 13, 2015 at 22:06

answered Dec 13, 2015 at 19:58

Dimitrios Bouzas

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4 Comments

Neil Kirk Over a year ago

Assuming that the size of a function pointer is not larger than the size of a data pointer. You should also add const to the cast to prevent accidents.

M.M Over a year ago

note: this conversion to void * is not required to exist

Dimitrios Bouzas Over a year ago

@M.M I know, mentioned the UB and proposed an alternative safer way.

Dimitrios Bouzas Over a year ago

@NeilKirk, mentioned the UB and proposed an alternative safer way.

juzzlin · Accepted Answer · 2015-12-13 20:09:21Z

0

You're not printing the address because it's now converted to a boolean value.

But you can do e.g. this:

std::cout << reinterpret_cast<unsigned long long int *>(func2) << std::endl;

Now you'll get the actual address.

edited Dec 13, 2015 at 20:09

answered Dec 13, 2015 at 20:01

juzzlin

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5 Comments

Neil Kirk Over a year ago

Assuming that the size of a function pointer is not larger than the size of unsigned long long int. You should also add const to the cast to prevent accidents.

juzzlin Over a year ago

Yeah, well...in C++11 long long is at least 64 bits.

Neil Kirk Over a year ago

So? What if pointers are 128-bits in the future? Once they were only 16-bits.

M.M Over a year ago

This may cause an alignment violation, if it even exists .. there's no reason to use any pointer type other than void *.

Neil Kirk Over a year ago

Sorry my mistake.. I didn't spot the *

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