Question

So I have this array in a header file like this:

// header.h
static const unsigned int array1[]={0x00,0x01,0x02,0x03};

And:

// file.c
main()
{
     unsigned int *ptrToArray;

     ptrArray = &array1[0];
 }

Correct me if I am wrong. I assume: to find the number of bytes of array elements, instead of sizeof(array1) the equivalent will be sizeof(*ptrArray), right?

And to access the elements of the array, instead of array[i], it will now be:

• *(ptrArray) for the first element,
• *(ptrArray+1) for the 2nd element so on right?

Answer 1

The type of *ptrToArray is int, therefore sizeof(*ptrToArray) is the same as sizeof(int). So it won’t tell you anything about the number of elements in array1.

Whilst you can write *(ptrArray+1), etc., you should just write ptrToArray[1]!