when do i use "&" within scanf()?

Question

This is my code:

int main() {
    int i,n, *arr;
    scanf("%d", &n);
    arr = (int *)malloc(sizeof(int)*n);
    for(i = 0; i < n; i++)
        scanf("%d", &arr[i]);
    printf("%lld", sumOfArray(arr,n));

    return 0;
}

Why do I have to use &arr[i]? I thought that it's not needed when arr is of type int *.

Simply put, because `arr[i]` is where you want the value to be stored, so you need to pass a *pointer* to that place. You could have used `arr+i` too (because `arr` decays into a pointer type). Also, [don't cast the result of `malloc(3)`](http://stackoverflow.com/a/605858/2793118) — Filipe Gonçalves, Aug 31 '15 at 20:51
There is no need to use this. If you want to scan `int hello`, you would write `&hello`. And you can use (almost) any name you want for your pointer. — too honest for this site, Aug 31 '15 at 21:04
@FilipeGonçalves: `arr` actually is a pointer type. (yes, I was also missleaded by the name) — too honest for this site, Aug 31 '15 at 21:31

score 1 · Answer 1 · answered Aug 31 '15 at 20:53

1

scanf() takes a memory address, which it resolves to a pointer to push data to; arr on its own would be a pointer, but arr[i] is an int. Therefore, you need to pass the address of arr[i] to scanf()

answered Aug 31 '15 at 20:53

F. Stephen Q

4,208
1
19
42

score 1 · Answer 2 · answered Aug 31 '15 at 20:53

1

Although arr is of type int *, that's not where scanf is putting the data. It's putting it in arr[i], which is of type int. Therefore, you need to pass the address of that variable, i.e. &arr[i].

answered Aug 31 '15 at 20:53

dbush

205,898
23
218
273

chux - Reinstate Monica · Answer 3 · 2015-09-01T16:26:35.827

Use & to get the address of an object.

Which of these makes sense?

scanf("%d", ... needs an address of an int as the next argument.

&n is an int *. Good

int n;
scanf("%d", &n);

arr is an int *. Good.

int *arr = malloc(1 * sizeof(*arr));
scanf("%d", arr);

arr[i] is an int: Bad

int *arr = malloc(n * sizeof(*arr));
for(i = 0; i < n; i++)
    scanf("%d", arr[i]); // bad

&arr[i] is an int *: Good

int *arr = malloc(n * sizeof(*arr));
for(i = 0; i < n; i++)
    scanf("%d", &arr[i]);

&arr[i] is an int *: Good

int arr[5];
for(i = 0; i < 5; i++)
    scanf("%d", &arr[i]);

arr is an "array 1 of int". Because it is an array, when passed to a function, arr is converted to the address of its first element. This is an int *: Good

int arr[1];
scanf("%d", arr);

&arr is the address of an "array 1 of int". Although this has the same value as &arr[0], it is not an int *: Bad

int arr[1];
scanf("%d", &arr); // Bad

Result scanf("%d", ... needs an int * argument.

When the modifiable object being passed converts to an int, an & is needed.
When the object being passed converts to an int *, no & is needed.

when do i use "&" within scanf()?

3 Answers3