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 converted the function pointer to `void*`, so does `printf` with `%p`, such as: ``` #include <iostream> using namespace std; int foo() {return 0;} int main() { int (*pf)(); pf = foo; cout << "cout << pf is " << pf << endl; cout << "cout << (void *)pf is " << (void *)pf << endl; printf("printf("%%p", pf) is %p
", pf); return 0; } ``` I compiled it with g++ and got results like this: ``` cout << pf is 1 cout << (void *)pf is 0x100000b0c printf("%p", pf) is 0x100000b0c ``` So, what does `cout` do with type `int (*)()`? I was told that the function pointer is treated as `bool`, is it true? And, what does `cout` do with type `void*`? EDIT: Anyhow, we can observe the content of a function pointer by converting it into `void*` and printing it out using `cout`. But, it does not work for member function pointers, and the compiler complains about an illegal conversion. I know that member fu…

Accepted Answer

There actually is an overload of the << operator that looks something like:

```
ostream & operator <<( ostream &, const void * );
```

which does what you expect - outputs in hex. There can be no such standard library overload for function pointers, because there are infinite number of types of them. So the pointer gets converted to another type, which in this case seems to be a bool - I can't offhand remember the rules for this.

Edit: The C++ Standard specifies:

4.12 Boolean conversions

1 An rvalue of arithmetic, enumeration, pointer, or pointer to member type can be converted to an rvalue of type bool.

This is the only conversion specified for function pointers.

How to print function pointers with cout?

Solution

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