Heya, so I can't wrap my mind a typedef'ed function pointer in a header I want to understand. It's of the form:
typedef void (*T_FUNC_CALLBACK) (UINT8 *setupPtr, UINT8 *dataPtr);
Tell me if I'm wrong: the type is void, and the name is T_FUNC_CALLBACK. But how can a function pointer be a void type? Am I missing something?
The type is "pointer to function returning void taking two arguments, each of type UINT8*".
I'm not too familiar with function ptr syntax, so I'm trying to understand: what does this typedef do for the code? Without it, I can already use *T_FUNC_CALLBACK. Since it has the same name, and same function, the typedef seems extraneous code.
Extraneous? You'd rather type:Originally Posted by ngloosh
Everywhere instead of:Code:void (*) (UINT8 *setupPtr, UINT8 *dataPtr)
? Weird.Code:T_FUNC_CALLBACK
EDIT: And your statement that "You can already use T_FUNC_CALLBACK" is untrue, since this typedef is what defines that name.
Last edited by brewbuck; 06-05-2008 at 10:31 AM.
You can't use T_FUNC_CALLBACK without the typedef. What the typedef does is make the declaration
equivalent toCode:T_FUNC_CALLBACK some_function
If you ever try to do *T_FUNC_CALLBACK in your code, it won't compile, in much the same way that you can't use *char to dereference a string.Code:void (*some_function)(UINT8 *setupPtr, UINT8 *dataPtr)
cool, cool. Both of you make sense. Thanks.
A better way to do it IMO, is instead of:
...do...Code:typedef void (*T_FUNC_CALLBACK) (UINT8 *setupPtr, UINT8 *dataPtr); T_FUNC_CALLBACK ptr = &something; ptr();
It makes it more obvious it's a pointer.Code:typedef void (T_FUNC_CALLBACK) (UINT8 *setupPtr, UINT8 *dataPtr); T_FUNC_CALLBACK* ptr = &something; ptr();
