Why Does "cout a function without calling it (not f() but f;). Print 1 Always?"

In this code, the code tries to "call" a function named pr without using parentheses. However, this is not actually calling the function. Instead, it is passing the function pointer to the cout function. When the function pointer is implicitly converted to a bool value, it is evaluated as true. Since true is equivalent to 1 in C , the output is always 1.

To clarify, the following lines from the provided code are not invoking the pr function:

pr;
cout 
To truly call the pr function, you would need to use parentheses like pr().
This behavior stems from the fact that function pointers are implicitly convertible to bool. In C  11, it is possible to overload the operator