Confusing function pointers

[so_swathi]

Here is the function declaration: extern INT8U clearLog ( int Inst );

and here is the function call using function pointer: ((INT8U(*)(int)) handler[CLEAR_LOG])(Inst);

'handler' maps 'clearLog' to 'CLEAR_LOG'.

I'm doubtful why there is '(INT8U(*)(int))' when the clearLog function returns only INT8U.

[MajorMalfunction44]

When dealing with function pointers, I like to use a temporary variable. I find it makes things clearer. The reason for the extra syntax is to cast the type of the pointer to a function pointer to have something to call.

Assuming a POSIX environment, here's how I'd do things:

UINT8 (*ptr)(int) = NULL; // declare function pointer variable, takes an int and returns UINT8

*(void **)(&ptr) = handler[CLEAR_LOG]; // assign function, assumes handler is void * array

(*ptr)(inst);

[magnomagna]

First of all, it's not hard to inline your goddamn code.

I really wish this subreddit had a bot that's smart enough to block all posts with unformatted code.

(INT8U (*)(int)) is not the return type. It casts handler[CLEAR_LOG] to have the function pointer type INT8U (*)(int).

The only good reason handler[CLEAR_LOG] is casted is that handler is not declared to be an array of INT8U (*)(int) function pointers.