Thread: Size of a function pointer?

Well, putting function pointers in an array of char would be a rather silly thing to do, as it would make much more sense to have an array of function pointers - that way you don't get problems with things that could trip you up. Such as alignment of function pointers that is different from char alignment on certain machines.

However, if you insist of doing silly things, sizeof() will tell you what the size is of a function pointer. Generally it's 1, 2, 4, 8 or 16 bytes - but there's no guarantee for any of those numbers.

Can you explain a bit more why you want to put function pointers in a char-array - it may be that we can come up with a better solution?

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Mats

If you got a function pointer like

Code:

void (*printhello)(void);

then the size of it is

Code:

sizeof(printhello)

.
If you have multiple function pointers than the array should be the size of the biggest function pointer.

Originally Posted by itCbitC

The sizeof operator can tell you that and it would be either 4 or 8 bytes depending on whether you're using the ILP32 or LP64 model.

Correct, however, the range of the answer is not limited to 4 or 8 - there are several other possible options - for example we could have a 32-bit segmented X86 model (I know of one compiler that can/could produce such code), in which case a function pointer is 6 bytes (but possibly aligned to 8 bytes), just as an example.

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Mats