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arrayscpointers
Zephram
Zephram

Reputation: 499

In C - Can I initialize an array (and set the size) from within a function?

I am trying to pass a pointer to a funcion, and in that function create an array, and set the passed pointer to that array.

What I'm trying to do is create an array inside the function and then use that array outside the funcion.

void createArray(int *myPointer)
{
    int intArray[20];

    *myPointer = &intArray[0]; // This line gets error:
                               // assignment makes integer from pointer without a cast [-Wint-conversion]

    for (int i = 0; i < 20; i++)
    {
        intArray[i] = i + 10;
    }
}

int main(void)

{
    int *intPointer;
    createArray(&intPointer);
    for (int i = 0; i < 20; i++)
    {
        printf("%d : %d \n", i, intPointer[i]);
    }
    return 0;
}

Upvotes: 1

Views: 174

Answers (2)

Lundin
Lundin

Reputation: 213892

The reason for the compiler error is the *myPointer dereferencing, which gives you the content of type int that the pointer points at. The compiler is saying that you can't assign the pointer &intArray[0] to a plain int.

Besides being invalid C, it wouldn't change where the pointer points. To do that, you would have to do myPointer = &intArray[0];. However, in this case you can't do that either because the myPointer is just a local pointer variable inside the function - a local copy of the pointer you passed. You need a way to update the pointer in main().

Furthermore, we can never return a pointer to local data (data with "automatic storage duration") because that data ceases to be valid as soon as we leave the function.

The normal solution is to create dynamically allocated memory instead, since such memory persists throughout the execution of the program (or until you explicitly free it). There are two different ways to write a function for that:

#include <stdlib.h> // for malloc

int* createArray (void)
{
    int* result = malloc(20 * sizeof(*result));
    if(result == NULL)
        return NULL; // some error handling will be required here

    for (int i = 0; i < 20; i++)
    {
        result[i] = i + 10;
    }
  
    return result;
}

...
int *intPointer = createArray();
...
free(intPointer);

Alternatively you could pass the pointer itself by value, so that by doing the *myPointer = ...; assignment, you refer to where a int** points at:

#include <stdlib.h> // for malloc

void createArray (int** myPointer)
{
    int* result;
    result = malloc(20 * sizeof(*result));
    if(result == NULL)
    { 
        /* some error handling will be required here */ 
    }

    for (int i = 0; i < 20; i++)
    {
        result[i] = i + 10;
    }
  
    *myPointer = result;
}

...
int *intPointer;
createArray(&intPointer);
...
free(intPointer);

Either of these are ok and which one you use is mostly a matter of design. It's easier syntax to have the pointer returned, but then normally the return of functions is reserved for some error code.

Upvotes: 1

ikegami
ikegami

Reputation: 385897

You meant to use

void createArray(int **myPointer)

That will resolve the two type errors.

But that will make main's pointer point to a variable that no longer exists after createArray returns. That's bad. And that's why you have to use malloc instead of automatic storage.

void createArray( int **myPointer ) {
    int *intArray = malloc( 20 * sizeof(int) );

    *myPointer = intArray;  // Same as `&intArray[0]`
    if ( !intArray )
       return;

    for ( int i = 0; i < 20; i++ )
        intArray[i] = i + 10;
}

int main(void) {
    int *intPointer;
    createArray(&intPointer);
    if ( intPointer ) {
       perror( NULL );
       exit( 1 );
    }

    for (int i = 0; i < 20; i++)
        printf( "%d: %d\n", i, intPointer[i] );

    return 0;
}

Upvotes: 3

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