Pointer to an array in Objective-C function call

Question

I want to pass an array of long long to an objective c function. I would like the array to be passed as a pointer, if possible, so that copying is not done. I wrote this code:

+ (int) getIndexOfLongLongValue: (long long) value inArray: (long long []) array size: (int) count
{
    for (int i =0; i < count; i++)
    {
       if (array[i] == value)
           return i + 1;
    }
    return 0;
}

to which I pass

long long varname[count];

as the second argument. I am wondering if I can pass this array as a pointer or if this method is fine. I don't need pointers to long longs but pointers to the array.

score 2 · Answer 1 · answered Jul 14 '12 at 15:09

2

C arrays always decay to pointers when passed to a function or a method, so it is not copied in your example. A pointer to long long is automatically a pointer to an array of long long, by virtue of the operator [] being applicable to all pointers. Since your method does not modify the array, you may want to add const to highlight this fact in your API; other than that, your method is fine.

answered Jul 14 '12 at 15:09

Sergey Kalinichenko

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Could you expand on what "decaying" means here. Decaying from what? What does an array define what a pointer doesn't? – Michel Müller Jul 14 '12 at 16:14
I think he means here by decay that the type becomes just a pointer whereas before is was a pointer that was called an array. – user1122069 Jul 14 '12 at 16:38
@MichelMüller See [this question](http://stackoverflow.com/q/1461432/335858) for an explanation. – Sergey Kalinichenko Jul 14 '12 at 18:19

score 2 · Accepted Answer · answered Jul 14 '12 at 15:10

2

It is being passed in as a pointer (language pedantry notwithstanding); nothing's getting copied. type[] is, for the most part, the same thing as type*.

To confirm this, check out sizeof(array) in the method. You'll see it's the same size as sizeof(void*).

If you really want a pointer to the array, i.e. a pointer to a pointer to the long longs, you'll need to use something like type**; but the only reason to want to do this is if you want to modify the underlying array pointer from the method, which is hardly ever the case.

answered Jul 14 '12 at 15:10

David Given

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Thanks, but I still don't understand fully. Lets say I have an array of 1000 items (as an example) and I pass it into this function. Does the memory used increase only by the size of one pointer - 4-8 bytes? – user1122069 Jul 14 '12 at 16:36
(long long *)array is the same as (long long []) array? – user1122069 Jul 14 '12 at 16:40
The array lives on the heap (usually). When you call the function, all you pass in is a pointer --- the location of the array. (It's actually quite hard to pass an array by value in C/C++/Objective C; you won't do it by accident.) Both `long long*` and `long long[]`, when used as parameter types, represent pointers. – David Given Jul 14 '12 at 19:14

score 0 · Answer 3 · answered Jul 14 '12 at 15:09

Objective-C is just an extremely thin wrapper around regular C, so this is handled the same as it would be handled in C: passing by reference.

+ (int)getIndexOfLongLongValue:(long long)value inArray:(long long *)array size:(int)count
{
    for (int i = 0; i < count; i++)
    {
        if (array[i] == value)
            return i + 1;
    }
}

Then in the code that calls the function: int myIndex = [object getIndexOfLongLongValue: aValue inArray: varname size: count];

Pointer to an array in Objective-C function call

3 Answers3