Printing a Function Pointer Without Invocation: Understanding the Mysterious Output of 1

In an intriguing code snippet, a function is "called" without invoking its parentheses ("pr;") and then printed using std::cout. Surprisingly, the result consistently yields 1, defying expectations.

The Enigma of Constant 1s

The code calls the function pr three different ways:

cout 
Intuition suggests that a function pointer should be printed instead of the enigmatic 1s. However, understanding this behavior requires delving deeper into C  's type conversion mechanisms.
Type Conversion and Bool Values
When passed as an argument to cout, pr is implicitly converted to a bool. This conversion occurs because bool is a fundamental type, and cout requires a fundamental type as input. In C  , a converted bool value is true if the original value is nonzero and false if it is zero.
Since pr is a function pointer, its value is a non-zero memory address. Consequently, when converted to bool, it evaluates to true, which is output as 1 by cout.
Customizing Function Pointer Printing (C  11 onwards)
C  11 introduces a customizable overload that allows for more informative printing of function pointers:
template 
std::ostream & operator
This overload prints the function pointer's address and the number of arguments it takes. It can be used to print pr as follows:
cout 
This approach provides more descriptive output, making it easier to understand the function pointer's properties.