Init value for a function pointer

Question

I have a simple question: how to give a function pointer a initial value, of kind of void..

typedef void (*FunctionPointer)();
FunctionPointer func; //how to init here
..... (func var may or may not be assigned a value here)
func()

So I want variable func to have an initial value, something like func = void() How can I achieve this? Thanks.

that would of course work but I dont want to define a blank function for this simple purpose. I just want to have a neat solution. — Maluvel, Apr 29 '14 at 06:57
Guess you are on your own then... unless you can find such a function that has no side-effects. — leppie, Apr 29 '14 at 06:59
Please do not hide away the pointer property in the typedef. Use `typedef void Function();` and `Function *func;` — hlovdal, Apr 29 '14 at 07:14
What do you want to achieve ? Maybe it's an [XY problem](http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem). — Jabberwocky, Apr 29 '14 at 07:18
Thanks all for your answers, yes it is XY problem (I read the def):D. I guess the only way is to define a blank function then. — Maluvel, Apr 29 '14 at 07:37
@hlovdal can you explain why it is not good to hide pointer property, currently I can: `func = CustomFunc;` but with your way I should do `func = &CustomFunc;` — Maluvel, Apr 29 '14 at 07:42
It should crash - this is the way to fix bugs. It looks like you want your program to continue working with uninitialized pointer. — Alex F, Apr 29 '14 at 07:43
See the question [http://stackoverflow.com/q/750178/23118](Typedef pointers a good idea?), and the answers are no. — hlovdal, Apr 29 '14 at 08:16

score 0 · Answer 1 · answered Apr 29 '14 at 07:04

0

Variables declared outside of any function already get initialized, you don't have to do anything.

If it is inside a function, then

FunctionPointer func = 0;

answered Apr 29 '14 at 07:04

M.M

1 Answers1