Why does printing a function name returns a value?

Question

I accidentally printed a function name without the parenthesis and it printed a value. I am just curious about how this happens? Output is same irrespective of the function

Answer 1

A mention of a function name without parentheses is interpreted as a function pointer. The pointer is interpreted as an unsigned int by the %u format specifier, which is undefined behaviour but happens to work on most systems.

The reason int k = foo doesn't work is that a function pointer normally needs a cast to be converted to int. However, printf is much more lenient because it uses varargs to parse its argument string; the type of the arguments is assumed to match the type requested in the format string.

Answer 2

std::cout << foo;

Outputs 1 because foo is a function pointer, it will be converted to bool with std::cout.

To print its address, you need explicitly cast it:

std::cout << reinterpret_cast<void*>(foo) << std::endl;

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With printf("\n%u", foo);, %u expects unsigned int, what you saw is the value of the function pointer converted to unsgigned int.

Answer 3

This statement printf("\n%u", foo); passes the address of function foo() to printf. So the value that gets printed is the address of this function in the memory of the running program.