CODEWITHSUNDEEP
c++functionsignature
codekiddy
codekiddy

Reputation: 6137

Is there a way to check function signature before calling it?

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);

Upvotes: 1

Views: 460

Answers (2)

Software_Designer
Software_Designer

Reputation: 8587

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)

Upvotes: 1

James McNellis
James McNellis

Reputation: 355079

The type of a function is known at compile-time. You can compare arbitrary types using is_same:

#include <iostream>
#include <type_traits>

int main()
{
    typedef void(*F0)(int);
    typedef void(*F1)(int, int);

    std::cout << std::is_same<F0, F0>::value << std::endl;
    std::cout << std::is_same<F0, F1>::value << std::endl;
}

Result:

1
0

The type trait value is a compile-time constant and can be used in template instantiation and for SFINAE.

Upvotes: 3

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