value of function in c/C++ [duplicate]

Question

Possible Duplicate:
How does dereferencing of a function pointer happen?

If we have

void f() {
    printf("called");
}

Then the following code will result in output of "calledcalled":

f();
(*f)();

I don't really understand how this works…what is the difference between *f and f? And why would you call a function using the latter syntax?

Answer 1

There are two ways to call a function in C++:

By name:

f();

Through a function pointer:

typedef void (*fptr_t)();
fptr_t fptr = &f;
(*fptr)();

Now, using the address-of operator on a function name (&f) obviously creates a function pointer. But the function name can implicitly convert to a function pointer when certain conditions are met. So the above code could be written as:

typedef void (*fptr_t)();
fptr_t fptr = f; // no address-of operator, implicit conversion
(*fptr)();

Your second example is doing exactly this, but using a temporary variable to hold the function pointer instead of a named local variable.

I prefer to use address-of when creating function pointers, the meaning is much clearer.

A related note: The function call operator will automatically dereference a function pointer if one is provided. So this also is legal:

typedef void (*fptr_t)();
fptr_t fptr = &f;
fptr();

That's pretty useful with templates, because the same syntax works whether you have a function pointer or a functor (object implementing operator()) passed in.

And neither shortcut works with pointer-to-members, there you NEED explicit address-of and dereference operators.

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In C, @Mehrdad explains that all function calls use a function pointer.