typedef with functions

[divine fallacy]

How would I define a function declared using typedef?

#include <stdio.h>

int main (int argc, char * argv)
{

  typedef int multiply(int arg1, int arg2);
  multiply mult_function_1;

  return 0;

}

I.e., provide the function body to mult_function_1; this does not work:

  multiply mult_function_1
  {
    int product = arg1 * arg2;
    return (product);
  }

[vlad from moscow]

According to the C++ Standard

A typedef of function type may be used to declare a function but shall not be used to define a function

So your code should look like

#include <stdio.h>

int main (int argc, char * argv)
{

  typedef int multiply(int arg1, int arg2);
  multiply mult_function_1;

  printf( "%d\n", mult_function_1( 2, 3 ) );

  return 0;
}

int mult_function_1( int x, int y )
{
   // for example
   return ( x * y );
}

[divine fallacy]

Thanks, but then typedef functions appear to be completely pointless. When
would one use them instead of typedef pointer functions?

[IceThatJaw]

typedef is useful for simple variable naming. It's also useful for renaming iterators. I can't recommend using it for functions/methods. That is just making whoever is reading the code more confused.

[Stewbond]

You can typedef a function pointer:

#include <cstdio>

int mult_function_1( int x, int y );

typedef int (*func)(int, int); // func is a function pointer to a function that accepts two ints and returns an int

int main (int argc, char * argv)
{
  func multiply = mult_function_1;

  printf( "%d\n", multiply( 2, 3 ) );

  return 0;
}

int mult_function_1( int x, int y )
{
   // for example
   return ( x * y );
}