How to fill array in c? (assignment makes integer from pointer without a cast)

Question

I am trying to fill a multidimensional array with chars.

The task is: I need to fill 30 names in array, each name could not be longer than 20.

char names_arr[NAME_BUF_SIZE][MAX_NAME_LENGTH];

So, I wrote such a method:

void read_from_input(char arr[][MAX_NAME_LENGTH])
{
    unsigned r = 0;
    int i = 0, j = 0;
    char tmp[MAX_NAME_LENGTH];

    for (i = 0; i < NAME_BUF_SIZE /*30*/; i++)
    {
        for (j = 0; j < MAX_NAME_LENGTH /*20*/; j++)
        {
            if((r = scanf("%s", tmp) != EOF) && (r != 0))
            {
                arr[i][j] = tmp;
            }
        }
    }
}

Such a way I would like to fill my array with names.

But the issue is that I get a such warning

main.c: In function ‘read_from_input’:
main.c:31:27: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
                 arr[i][j] = tmp;
                           ^

What am I doing wrong?

What is the purpose of tmp array? Here arr[i][j] = tmp; you're trying yo assign the array to the element of second array. — funnydman
– funnydman, Commented Dec 21, 2019 at 12:56
arr[i][j] is a char, and you are assigning to it tmp that is a char array, that is actually a pointer to the first char of the array. — Roberto Caboni
– Roberto Caboni, Commented Dec 21, 2019 at 12:58
Also, in real application, if you use scanf, always limit the size of the input to avoid buffer overruns (e.g.: "%19s"). — dash-o
– dash-o, Commented Dec 21, 2019 at 13:25
Aleksey, strcpy is ok. And about the suggestion above, an array of 20 chars can contain a string long up to 19chars because uone of them is the stringo terminator '\0'. — Roberto Caboni
– Roberto Caboni, Commented Dec 21, 2019 at 14:27

Saucy Goat · Accepted Answer · 2019-12-21 16:51:02Z

1

Perhaps this is what you mean to do:

for (i = 0; i < NAME_BUF_SIZE; i++) {
    if((r = scanf("%19s", tmp) != EOF) && (r != 0)) {
        strcpy(arr[i], tmp);
    }
}

First of all, in your original code, you're trying to assign tmp (char *) to arr[i][j] (char). If arr is an array of words, then each a[i] will be a word (char*) and each a[i][j] will be a letter (char). Not only that, but that is not the correct way to copy a string.

This is how you should copy a string in C:

strcpy(arr[i], tmp);

Hope this helped!

answered Dec 21, 2019 at 16:51

Saucy Goat

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