format specifies type 'int' but the argument has type 'int *'

Question

I get this error: format specifies type 'int' but the argument has type 'int *' [-Werror,-Wformat]

#include <stdio.h>

int main(void)
{
    int arr[5] = {0};

    for(int i = 1; i < 5; i++)
    {
        arr[i] = i + 1;
    }
    printf("%i", arr);
}

`arr` is an array, which decays to a pointer `int*` when passed to a function. It isn't the `int` expected by the format specifier `%i`. — Weather Vane, Feb 18 '22 at 22:19
There's no "print all" format specifier for an array. If you want to print the array, you need to `printf` in a loop. — yano, Feb 18 '22 at 22:20
you cant print a whole array with one printf if thats what you are trying to do, you have to loop — pm100, Feb 18 '22 at 22:21
Aside: notice that `arr[0]` has not been written to, since the initialisation, unlike all the other array elements. — Weather Vane, Feb 18 '22 at 22:22
You may want to read this: [What is array to pointer decay?](https://stackoverflow.com/q/1461432/12149471) — Andreas Wenzel, Feb 18 '22 at 22:27

score 4 · Accepted Answer · edited Feb 18 '22 at 22:41

In C, whenever you pass an array to a function, only a pointer to the first element of said array is passed into the function (this is called "array to pointer decay"). In order to print every element inside an array, you have to access every single one of them and print them individually, using the [] operator. Like this:

for(int i = 0; i < 5; i++){
    printf("%i", arr[i]);
}

Also, remember that arrays are 0 based, meaning arr[0] is the first element, but arr[5] is not a valid element, as it's out of bounds.

format specifies type 'int' but the argument has type 'int *'

1 Answers1