Using scanf to fill out a multidimensional array (C language)

Question

I was wondering how to properly use scanf to fill out a multidimensional array.

Here's my code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, const char * argv[]) {

int n; //number of rounds
int* sArray; //multidimensional array that holds the scores of both players
int i;

scanf("%d", &n);

sArray = (int*) calloc (n * 2, sizeof(int));


for(i=0; i<n; i++) {
    scanf("%d %d", &sArray[i][1], &sArray[i][2]); 

}

return 0;
}

It gives me an error, "Subscripted value is not an array, pointer, or vector." Any help would be much appreciated!

To dereference a two dimensional array you need a double pointer: int** sArray; — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Jan 17, 2016 at 23:14
Just FYI, your #include statements should be before your main method. — AlexPogue
– AlexPogue, Commented Jan 17, 2016 at 23:23
Always check the function return value from scanf (the number of items successfully scanned). — Weather Vane
– Weather Vane, Commented Jan 17, 2016 at 23:26
To fill a two dimensional array you need two nested for loops — Claudio Cortese
– Claudio Cortese, Commented Jan 17, 2016 at 23:29
Should be int (*sArray)[2] = calloc (n , sizeof **sArray * 2); — David C. Rankin
– David C. Rankin, Commented Jan 17, 2016 at 23:36

neshkeev · Accepted Answer · 2016-01-17 23:56:00Z

2

A two dimentional array is defined as follows: int sArray[N][M], but since you wanted to work with the dynamic memory I offer you to take a look at a pointer to pointer at int:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n;
    scanf("%d", &n);
    int **sArray;
    sArray = (int **)malloc(n * sizeof(int *));

    int i;
    for(i = 0; i < n; i++)
    {
        sArray[i] = (int *)malloc(2 * sizeof(int));
        scanf("%d %d", &sArray[i][1], &sArray[i][2]);
    } 
    return 0;
}

Don't forget to clean-up after you are done with the array.

As mentioned in the commentaries, You don't need to cast the result of malloc if you work with pure c. I did this because my c++ compiler refused to compile it without this cast.

You might need to check errors during a dynamic allocation of the array. Read more here

edited Jan 17, 2016 at 23:56

answered Jan 17, 2016 at 23:30

neshkeev

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9 Comments

user4520 Over a year ago

That's fine, but please don't cast the result of malloc.

hyde Over a year ago

That's not really a two-dimensional array, that is 1-dimensional array of pointers to 1-dimensional arrays (with massive memory overhead in this case). It is also possible to have actual 2-dimensional array of values.

user4520 Over a year ago

@hyde Not if the C implementation you're using doesn't have VLAs, which is possible in C11.

hyde Over a year ago

@szczurcio You don't need VLA if other dimension is fixed at compile time, and here it is 2. So C89 is enough. Also, C99 already has VLA.

ryyker Over a year ago

A two dimentional array is defined as follows: int **sArray... A two dimensional array in C looks like: int array[x][y]; What you have is a pointer to pointer to int: int **array; that will be used to emulate a 2D array.

|

Christophe · Accepted Answer · 2016-01-18 00:04:20Z

There are already a lot of good answers here on how to define your dynamic 2D array. But this variant was not yet mentionned, so I put it for the records.

As the last dimension of your array is fixed, you could define your 2D array as follows:

int (*sArray)[2]; //multidimensional array that holds the scores of both players
...
sArray = (int(*)[2]) calloc (n, sizeof(int)*2);  // self explaining

In this way, all the elements will be stored contiguously (each n element of the allocated array, is 2 contiguous integers), without the need for an array to arrays.

The rest of your code remains identical. Except that you shoud address sArray[i][0] and ..[1] instead of [1] and [2] and free memory at the end. In C array indexing starts always from 0 and goes to size-1.

Of course, this approach is strictly limited to 2D arrays where the last dimension is fixed.

Live demo with addressing

Claudio Cortese · Accepted Answer · 2016-01-17 23:33:03Z

0

Usually to fill a bidimensional array you will use two nested for loops. For example :

int array[2][3] = {0};

for (i = 0; i < 2; i++)
   for (k = 0; k < 3; k++)
      scanf("%d", &array [i][k]);

answered Jan 17, 2016 at 23:33

Claudio Cortese

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Comments

user3813674 · Accepted Answer · 2016-01-17 23:59:09Z

0

You could do this too:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, const char * argv[]) {


    int n; //number of rounds
    int** sArray = malloc(2 * sizeof(int*)); //multidimensional array that holds the scores of both players
    scanf("%d", &n);
    sArray[0] = (int*)calloc(n , sizeof(int));
    sArray[1] = (int*)calloc(n , sizeof(int));

    int i;
    for (i = 0; i < n; i++) {
        scanf("%d %d", &sArray[0][i], &sArray[1][i]);
    }

    free(sArray[0]);
    free(sArray[1]);
    free(sArray);

    return 0;
}

edited Jan 17, 2016 at 23:59

answered Jan 17, 2016 at 23:37

user3813674

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3 Comments

David C. Rankin Over a year ago

That is somewhat of a painful way to allocate that does not quite accomplish what you have set out to do. Look closely at the comment for how to allocate a pointer-to-array 2 of type int or at the other answer to allocating a pointer-to-pointer-to type int.

user3813674 Over a year ago

@DavidC.Rankin: Got it!

user4520 Over a year ago

Uh, this is UB. You want to read n before passing it to calloc.

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