Question

I want to create an integer pointer p, allocate memory for a 10-element array, and then fill each element with the value of 5. Here’s my code:

//Allocate memory for a 10-element integer array.
int array[10];
int *p = (int *)malloc( sizeof(array) );

//Fill each element with the value of 5.
int i = 0;
printf("Size of array: %d\n", sizeof(array));
while (i < sizeof(array)){
    *p = 5;
             printf("Current value of array: %p\n", *p);
    *p += sizeof(int);
    i += sizeof(int);
}

I’ve added some print statements around this code, but I’m not sure if it’s actually filling each element with the value of 5.

So, is my code working correctly? Thanks for your time.

Answer 1

Here’s a more idiomatic way of doing things:

/* Just allocate the array into your pointer */
int arraySize = 10;
int *p = malloc(sizeof(int) * arraySize);

printf("Size of array: %d\n", arraySize);

/* Use a for loop to iterate over the array */
int i;
for (i = 0; i < arraySize; ++i)
{
    p[i] = 5;
    printf("Value of index %d in the array: %d\n", i, p[i]);
}

Note that you need to keep track of your array size separately, either in a variable (as I have done) or a macro (#define statement) or just with the integer literal. Using the integer literal is error-prone, however, because if you need to change the array size later, you need to change more lines of code.