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A question about function pointer in C...

What does this do exacty?

Code:

int(*fp)(void) = (int(*)(void))buf

This declares a function pointer that takes a void argument and returns an int. However I don't understand what this assignment does exactly.

Quote:

Originally Posted by trist007 (Post 4429493)

What does this do exacty?

Code:

int(*fp)(void) = (int(*)(void))buf

This declares a function pointer that takes a void argument and returns an int. However I don't understand what this assignment does exactly.

int(*)() is a type and int(*fp)() is the way you use it.

Code:

typedef int(*f_pointer)();

f_pointer fp = (f_pointer) buf;

The assignment takes the address of the function so that *fp is a reference (in C++, anyway) to the function, which can be evaluated with ().

Code:

int value = (*fp)(); //normal function-pointer evaluation



int(&fp2)() = *fp; //also allowed in C++

value = fp2();

Kevin Barry

Ah ok I see, thank you.