Dynamically allocate an array of specified size in C

Question

I need to create an array that's size is determined by user input, and then has pointers to said array. All the array will hold is random numbers between 500-600. I can't seem to use malloc correctly. I am still new to C, so help is appreciated.

int main(){
        int size;
    printf("Enter size of array");
    scanf("%d", &size);


    int array[size];
    int *aPtr = (int *) malloc(sizeof(int) * array);

Please don't cast the return value of malloc in C - it can hide problems that you don't want hidden. C is perfectly capable of changing a void* to any other pointer implicitly. — paxdiablo
– paxdiablo, Commented Sep 23, 2012 at 21:13

Inisheer · Accepted Answer · 2012-09-23 20:52:05Z

5

You only need:

int *aptr = malloc(sizeof(int) * size);

and then you can access it just like an array.

aptr[0] = 123;

answered Sep 23, 2012 at 20:52

Inisheer

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Comments

LSerni · Accepted Answer · 2012-09-23 20:54:23Z

1

You probably wanted to write:

int *aPtr = (int *) malloc(sizeof(int) * size);

You don't need that array variable anyway. You can index aPtr like aPtr[10]. Also don't forget free(aPtr) at the end.

edited Sep 23, 2012 at 20:54

LSerni

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answered Sep 23, 2012 at 20:52

petersohn

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