Dynamically allocate memory for an int array converted to string in C

Question

I have an integer array that is dynamically allocated by the following:

int *ranArr(int size) {
    int *arr= malloc(size *sizeof(int));

    for(int i = 0; i < size; i++) {
        arr[i] = genRan(size);
    }

    return arr;
}

I want to dynamically allocate memory for a string that I will convert my integer array into and store it in the string:

        int *arrayCopy = ranArr(size);
        int arrayCopyLen = size;
        char *arrayString= (char*) malloc(size * sizeof(char));
        int index = 0;
        for(int i = 0; i< arrayCopyLen ; i++) {
            index += sprintf(&arrayString[index], "%d ,", arrayCopy[i]);
        }
        int arrayStringLen = strlen(arrayString)+1;

This works for array sizes around 1000 but when I set the size to 2000 or so it breaks and I get a segmentation fault.

@EugeneSh. This function takes an integer as parameter (typeof size) and returns an integer (typeof arr[i]). This function calculates something that is unimportant for the problem. The function name suggests some random number generation, but that's a guess. — harper
– harper, Commented Jun 12, 2018 at 16:45
@harper that is correct, it just generates a random number from 0 to size. — xnv23
– xnv23, Commented Jun 12, 2018 at 17:13

dbush · Accepted Answer · 2018-06-12 16:40:14Z

3

You're not allocating enough memory for your string.

Converting a single int to a string takes up up to 12 bytes, assuming an int is 32 bit. For example, converting the value -2000000000 to a string needs 12 characters. So you need at least 12 * size bytes for your string, plus room for the ", " characters and the null terminator.

Also, sizeof(char) is 1 by definition, so no need to multiply by that.

char *arrayString = malloc(size * 14 + 1);   
// 14 = 12 bytes for a 32-bit int plus 2 bytes for ", "

answered Jun 12, 2018 at 16:40

dbush

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