pointer array when using scanf()

Question

In my code, I used pointer array like this.

arr = (int*)malloc(N * sizeof(int));

        for (i = 0; i < N; i++)
            scanf("%d", (arr + i));

Because pointer arithmetic points address.

But I wonder if it's right to only write (arr+i) without * or &.

What is right?

Please help me.

What you have is correct. The scanf arg needs to be an int * which is exactly what you have. Using * would be wrong as that would give an int and using & would be wrong as that would give an int **. — kaylum
– kaylum, Commented Sep 24, 2020 at 11:47
For any pointer or array a and index i, the expression *(a + i) is exactly equal to a[i]. From that follows that &a[i] is equal to &*(a + i) which is equal to a + i. Which is exactly what you have. — Some programmer dude
– Some programmer dude, Commented Sep 24, 2020 at 11:49

Vlad from Moscow · Accepted Answer · 2020-09-24 11:56:41Z

6

The expression arr + i has the type int * and is equivalent to the expression &arr[i] (or even to the expression &i[arr] :)).

So you may write either

scanf( "%d", &arr[i] );

or

scanf( "%d", arr + i );

answered Sep 24, 2020 at 11:48

Vlad from Moscow

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