Why Does Printing a Function Pointer Without Invocation Yield \'1\' in C ?

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:

<code class="cpp">cout << pr;  // output: 1
cout << *pr; // output: 1
cout << &pr; // output: 1

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:

<code class="cpp">template <class RType, class ... ArgTypes>
std::ostream & operator<<(std::ostream & s, RType(*func)(ArgTypes...))
{
    return s << "(func_ptr=" << (void*)func << ")(num_args=" << sizeof...(ArgTypes) << ")";
}</code>

This overload prints the function pointer's address and the number of arguments it takes. It can be used to print pr as follows:

<code class="cpp">cout << pr; // output: (func_ptr=0x12345678)(num_args=0)</code>

This approach provides more descriptive output, making it easier to understand the function pointer's properties.

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