What is the difference between declarations: void (allocators[])(size_t) and void allocators[] ?

Question

40 different allocation functions give 40 different call sites

void f00(size_t sz) { void* ptr = malloc(sz); free(ptr); }
void f01(size_t sz) { void* ptr = malloc(sz); free(ptr); }
...
void f39(size_t sz) { void* ptr = malloc(sz); free(ptr); }

An array of those allocation functions. How I need to define?

like this: why (size_t) ?:

  void (*allocators[])(size_t) = { &f00, &f01, ... , &f39 };

or

  void* allocators[] = { &f00, &f01, ... , &f39 };

and what the difference between this two declarations?

possible duplicate of [How define an array of function pointers in C](http://stackoverflow.com/questions/5488608/how-define-an-array-of-function-pointers-in-c) — juanchopanza, Apr 19 '15 at 16:47
Please see my previous answer: http://stackoverflow.com/questions/29310161/alternative-to-many-case-switch-statement-in-c/29310574#29310574 — Weather Vane, Apr 19 '15 at 16:48
Already answered, take a look at http://stackoverflow.com/questions/5309859/how-to-define-an-array-of-functions-in-c — Mohamed Amjad LASRI, Apr 19 '15 at 16:49
OP also asked *why `(size_t)` ?* Because that is the argument of the functions. — Weather Vane, Apr 19 '15 at 16:57
@WeatherVane this is a rule? Can I use the second declaration? — Anatoly, Apr 19 '15 at 16:59
@WeatherVane This is a different question because I asked why we need to put (size_t) and what the difference between two declarations — Anatoly, Apr 19 '15 at 17:02
Don't you think both of your questions have been answered here? — Weather Vane, Apr 19 '15 at 17:06

haccks · Accepted Answer · 2015-04-19T17:05:03.380

2

First is correct. In first declaration no need of & in intializers. It should be

 void (*allocators[])(size_t) = { f00, f01, ... , f39 };

It declares allocators as an array of pointers to function that return nothing and expects an argument of size_t type.
(size_t) informs the compiler that all functions expects an argument of this type.

Second is wrong because it declares an array of pointers to void.

edited Apr 19 '15 at 17:05

answered Apr 19 '15 at 16:48

haccks

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They are not both wrong. The version with `&f00, &f01, ...` is correct and is equivalent to your answer. You can argue that `&` is redundant, but that doesn't make it *wrong*. – Filipe Gonçalves Apr 19 '15 at 16:52
@FilipeGonçalves; Apologies. I was out of mind then. You are right. – haccks Apr 19 '15 at 16:56
Why (size_t) in the first variant? I mean void (*allocators[])(size_t) – Anatoly Apr 19 '15 at 16:57
1

The first is correct. The second is wrong (you switched it in your answer). – Filipe Gonçalves Apr 19 '15 at 16:59
@Anatoly Because that's what the functions expect as an argument. – Filipe Gonçalves Apr 19 '15 at 17:00
1

@Anatoly Because someone wrote them that way. If you want to declare a pointer to a function receiving `X`, then you'll have to specify `X` as part of the pointer declaration, don't you agree? – Filipe Gonçalves Apr 19 '15 at 17:01
2

Because it's the function argument you declared for every function in the array, and the compiler needs to know what arguments are needed when one of the functions in the array is called. – Weather Vane Apr 19 '15 at 17:02

What is the difference between declarations: void (*allocators[])(size_t) and void* allocators[] ?

1 Answers1

What is the difference between declarations: void (allocators[])(size_t) and void allocators[] ?