How to print function pointers with cout?

Question

I want to print out a function pointer using cout, and found it did not work. But it worked after I converting the function pointer to (void *), so does printf with %p, such

Answer 1

You can think of a function pointer as being the address of the first instruction in that function's machine code. Any pointer can be treated as a bool: 0 is false and everything else is true. As you observed, when cast to void * and given as an argument to the stream insertion operator (<<), the address is printed. (Viewed strictly, casting a pointer-to-function to void * is undefined.)

Without the cast, the story is a little complex. For matching overloaded functions ("overload resolution"), a C++ compiler gathers a set of candidate functions and from these candidates selects the "best viable" one, using implicit conversions if necessary. The wrinkle is the matching rules form a partial order, so multiple best-viable matches cause an ambiguity error.

In order of preference, the standard conversions (and of course there also user-defined and ellipsis conversions, not detailed) are

• exact match (i.e., no conversion necessary)
• promotion (e.g., int to float)
• other conversions

The last category includes boolean conversions, and any pointer type may be converted to bool: 0 (or NULL) is false and everything else is true. The latter shows up as 1 when passed to the stream insertion operator.

To get 0 instead, change your initialization to

pf = 0;

Remember that initializing a pointer with a zero-valued constant expression yields the null pointer.