Difference between *array and array[0] when calculating number of array elements in C

Question

For the given array:

int array[10];

What is the difference between

sizeof array / sizeof *array

and

sizeof array / sizeof array[0]

when calculating the number of elements in array?

Answer 1

There is none in this case. Both ultimately results in the type int because in the first case array decays into pointer to the first element when used as an operand of *(You are dereferencing a int*).

Note one thing - you might have heard that on sizeof there is no decay of arrays but here the decay happens when array is used as operand to *. The operand for sizeof is simply int. (Discussing about sizeof arr/sizeof *arr)

In case you want to get a clear idea - try this

int a[10][20];
printf("%zu %zu\n",sizeof a/ sizeof *a, sizeof a/ sizeof **a);

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Your second case is WYSIWYG type of thing - what I mean by this, it just says total size of the array and divide it with what it contains (the size of it). And yes it would return the size of the array. That is what is being done.

arr[0] is nothing but *(arr+0) means *arr. So that makes sense and they are same.

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Few points that will clarify things (If you later use sizeof)

_{array and &array[0] is same in this context. Their value and type both will be same. But For pointers being the same value is just one thing - their implication on the pointer arithmetic is a whole different story. int a[10][20]; Here a and &a and a[0] all has same value but their type is different and since pointer arithmetic is closely coupled with what it points to. It matters.}

Answer 2

The expression array[n] is by definition the same as *(array + n)

The expression array + 0 is by definition the same as array (unless array is an operand of sizeof).

So yes, array[0] and *array are the same.

This is much more than is needed in this case. For arguments of sizeof only the type matters, and it is certainly the same for both expressions.