Initializing a pointer in a separate function in C

Question

I need to do a simple thing, which I used to do many times in Java, but I'm stuck in C (pure C, not C++). The situation looks like this:

int *a;

void initArray( int *arr )
{
    arr = malloc( sizeof( int )  * SIZE );
}

int main()
{
    initArray( a );
    // a is NULL here! what to do?!
    return 0;
}

I have some "initializing" function, which SHOULD assign a given pointer to some allocated data (doesn't matter). How should I give a pointer to a function in order to this pointer will be modified, and then can be used further in the code (after that function call returns)?

score 27 · Accepted Answer · answered Mar 21 '10 at 06:41

27

You need to adjust the *a pointer, this means you need to pass a pointer to the *a. You do that like this:

int *a;

void initArray( int **arr )
{
    *arr = malloc( sizeof( int )  * SIZE );
}

int main()
{
    initArray( &a );
    return 0;
}

answered Mar 21 '10 at 06:41

Michael Anderson

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3

Alternatively (and better IMHO) `int *initArray(size_t s) { return malloc(sizeof(int) * s); }` and then use it as `int *arr = initArray(SIZE);` – Chris Lutz Mar 21 '10 at 06:43
1

Chris, I shouldn't, because this example is a kind of unnatural one. The "initArray" function in my real situation returns a value, so I can't use it. But thanx anyway :) – pechenie Mar 21 '10 at 06:47

score 8 · Answer 2 · answered Mar 21 '10 at 06:41

You are assigning arr by-value inside initArray, so any change to the value of arr will be invisible to the outside world. You need to pass arr by pointer:

void initArray(int** arr) {
  // perform null-check, etc.
  *arr = malloc(SIZE*sizeof(int));
}
...
initArray(&a);

Initializing a pointer in a separate function in C

2 Answers2

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