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

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(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.