function name in if statement is converted in a strange way

Question

With this code (valid C++11):

#include <stdio.h>
#include <typeinfo>

bool my_awesome_func(int param) {
  return (param > 1);
}

int main(int argc, char const *argv[]) {
  fprintf(stderr, "type of my_awesome_func: %s\n", 
          typeid(my_awesome_func).name());
  if (my_awesome_func) {
    fprintf(stderr, "WHAT???\n");
  }
  return 0;
}

The question is inside of the if statement. While typeid returns me something that looks like FbiE (which I think is gcc language for function type) I do not understand why this type is being implicitly converted into bool (just an example, also works with int).

Why does the if statement compile and evaluate true?

Answer 1

There is no casting in your code. A cast is an explicit conversion. I assume you are asking: What does implicit conversion of a function to bool do?

The answer to that is: the function is converted to a function pointer. Then the function pointer is converted to bool via implicit conversion. That conversion is defined as yielding:

• false for a null function pointer
• true for any other function pointer

So in your code, the body of if (my_awesome_func) is always entered. (Converting an actual function to a function pointer never yields a null pointer).