Function parameter as array with declared size

Question

I frequently use the following convention to inform client code that a function requires an argument of an array with defined size:

/* File foo.h */
int foo (int arg[10]);

The message I want to give to client code is that they must provide an array of type int with 10 positions.

I am aware that it is not very usual, so I came here to ask: Am I missing any side effect of this convention ? Is it anyhow harmful?

Thank!

Why not use `int foo (int arg[/*10*/]);`? Commenting the dimension — David Ranieri, Mar 10 '15 at 18:07
possible duplicate of [C Parameter Array Declarators](http://stackoverflow.com/questions/3693429/c-parameter-array-declarators) — Colonel Thirty Two, Mar 10 '15 at 18:09
@WeatherVane It's intended for humans, as a way to let them know that the array should always have 10 elements. — Barmar, Mar 10 '15 at 18:09
You can wrap the array to a struct and pass a pointer to that struct in. Arrays in structs have size info. — user3528438, Mar 10 '15 at 18:10
@user3528438 Would you tell me more about the difference between the two ? — Felipe Lavratti, Mar 10 '15 at 18:16
@Barmar I was thinking it could be enforced by error checking at runtime. — Weather Vane, Mar 10 '15 at 18:17

score 2 · Answer 1 · answered Mar 10 '15 at 18:20

2

struct arrayContainerTen{
  int data[10];
}
void aFunction(struct arrayContainerTen *pAnArray)
{
    size_t size = sizeof(pAnArray->data);
}
main()
{
     arrayContainerTen anArray;
     aFunction(&anArray);
}

answered Mar 10 '15 at 18:20

user3528438

2,737
2
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score 2 · Accepted Answer · answered Mar 10 '15 at 18:21

If you want to insist on getting an array of size 10, you can use:

int foo (int (*arg)[10]);

The ill-side effects of this are:

In the function, you have to use:
```
(*arg)[index] 
```
instead of just
```
arg[index]
```

The calling function must use:

int array[10];
foo(&array);

instead of

int array[10];
foo(array);

You cannot use an array that has more than 10 elements.
```
int array[20];
foo(&array);   // Not OK.
```

You cannot use a malloced array.

int* array = malloc(sizeof(int)*10);
foo(array);   // Not OK.

Now pick the solution that is least harmful.

At least, it's safe to cast the pointer to the array if it's too big or allocated: for 3. : `foo((void *)array)` or `foo((int(*)[])array)`; for 4.: `foo((int(*)[])array)` etc. — mafso, Mar 10 '15 at 18:40

score 1 · Answer 3 · answered Mar 10 '15 at 18:10

1

There's no harm in writing it like this. But just be aware that the compiler will not enforce the requirement. A declaration like that is treated by the compiler as if you'd written.

int foo(int *arg);

answered Mar 10 '15 at 18:10

Barmar

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Function parameter as array with declared size

3 Answers3

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