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Code edits, don't hide pointer

edited May 19, 2017 at 12:44

In situations like this, do a typedef to name your function pointersignature, that makes it far simpler.:

typedef void (*MESSAGE_HANDLER)MESSAGE_HANDLER(void);

with that in place, it should be just:

const MESSAGE_HANDLER * const handlers[] = { function1, function2 };

To get the actual content of the array constant.

EDIT: Removed pointer part from the typedef, this really is better (live and learn).

In situations like this, do a typedef to name your function pointer, that makes it far simpler.

typedef void (*MESSAGE_HANDLER)(void);

with that in place, it should be just:

const MESSAGE_HANDLER handlers[] = { function1, function2 };

To get the actual content of the array constant.

In situations like this, do a typedef to name your function signature, that makes it far simpler:

typedef void MESSAGE_HANDLER(void);

with that in place, it should be just:

MESSAGE_HANDLER * const handlers[] = { function1, function2 };

To get the actual content of the array constant.

EDIT: Removed pointer part from the typedef, this really is better (live and learn).

answered Dec 3, 2008 at 15:23

In situations like this, do a typedef to name your function pointer, that makes it far simpler.

typedef void (*MESSAGE_HANDLER)(void);

with that in place, it should be just:

const MESSAGE_HANDLER handlers[] = { function1, function2 };

To get the actual content of the array constant.