Question

for example this works:

if ( typeid( int) == typeid( int) ) //...

how to do the same with function signatures?

if (typeid (void (*)(void) ) == typeid( void(*)(void) ) //that of course dosn't work

how do we check thos two for signature?

void f(int);
int x(double);

Answer 1

Use typeid(foo).name() .

For instance : if ( typeid(func1).name() == typeid(func2).name() ) //do stuff

#include <cstdio>
#include <iostream>
#include <typeinfo>
using namespace std ;

void foo()
{    
}

int bar()
{   
    return 1;
}

int main(void)
{
   if (typeid(foo).name() == typeid(bar).name())
       cout<<typeid(foo).name()<<" equals "<<typeid(bar).name()<<" \n";
   else
   if (typeid(foo).name() != typeid(bar).name())
       cout<<typeid(foo).name()<<" is not equal to "<<typeid(bar).name()<<" \n";

   cout << "\nPress ENTER to continue \n\n";   cin.ignore();  // pause screen

   return 0;
}

output:

void (__cdecl*)(void) is not equal to int (__cdecl*)(void)