How Can I Print Function Pointers Using `cout` in C ?

Printing Function Pointers with cout

In C , cout does not directly support printing function pointers. However, using a specific operator overload allows it to be done by converting the function pointer to a void pointer (void*).

The operator<< is overloaded to work with void pointers as follows:

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

This overload allows the output of void pointers in hexadecimal format. Function pointers cannot have a standard library overload for operator<< because they exist in an infinite number of types.

When attempting to print a function pointer directly with cout, it is implicitly converted to another type. In many cases, this conversion results in a boolean value, but the exact conversion rules are implementation-defined.

To print a function pointer with cout, the following steps can be taken:

1. Declare and initialize a function pointer:

int (*pf)();
pf = foo;

2. Cast the function pointer to a void pointer:

cout << "cout << (void *)pf is " << (void *)pf << endl;

3. Use the overloaded operator<< to print the void pointer:

cout << "cout << (void *)pf is " << (void *)pf << endl;

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