User input from one line into multi dimention array

Question

So i have this example:

3 4 6 11 4 6 38 7 6 9

I need to get from user several numbers and create multi dimension array.

The first number N (3 in my example) mean that my array (or matrix) will contain N² value and insert all this numbers (the next 9 values) into my array.

So first i need to define the array according my first value (3 in my example) and create my array:

int a[3][3];

The catch here is that i need to get all my input in a single line so i cannot use this:

int marks[3];
int i;

for(i=0;i<3;i++)
{
    printf("Enter a no\n");
    scanf("%d",(marks+i));
}

Answer 1

scanf reads data from stdin and stores them according to the parameter format into the locations pointed by the additional arguments.

scanf will store whole the input from stdin, so it really does not matter if it is in single line or not. The way for your code is just, simply ask user for dimensions and create the variable of the given dimension and fill each value with a scanf.

There are 3 ways to do it :

1. If your compiler supports variable length array:

   int size,i,j;
   scanf("%d",&size);
   
   int matrix[size][size];
   
   for(i=0;i<size;i++)
   {
       for(j=0;j<size;j++)
       {
           scanf("%d",&matrix[i][j]);
       }
   }
   

2. Does not support variable length array: (using array of pointers)

   int size,i,j;
   scanf("%d",&size);
   
   int **matrix;
   matrix = (int **)malloc(sizeof(int *) * size);
   
   for(i=0;i<size;i++)
   {
       matrix[i] = (int *)malloc(sizeof(int) * size);
   }
   
   for(i=0;i<size;i++)
   {
       for(j=0;j<size;j++)
       {
           scanf("%d",&matrix[i][j]);
       }
   }
   

   This looks easy but the way is very poor as it is not actually an 2D array but an array of pointers. hence, I suggest to use the 3rd way if your compiler does not support variable length array.
   
   

3. Using malloc: (with pointer)

   Create Array of integers like:

   int size;
   scanf("%d",&size);
   int *matrix;
   matrix = (int *) malloc( sizeof(int) * size * size );
   

   Now, to fill or get the values of the Array you need to traverse to the particular value.

   to fill:

    for(i=0;i<size;i++)
    {
       for(j=0;j<size;j++)
       {
           scanf("%d", (matrix + i*(size) + j));
       }
    }
   

   to iterate: (print in this case)

    for(i=0;i<size;i++)
    {
       for(j=0;j<size;j++)
       {
           printf("%d " ,*(matrix+ i*size + j) );
       }
       printf("\n");
    }
   

Note: It is very important to constrain the input and check whether it lies in some fixed range or not. for ex: ( 0 < size < 100). It is very simple and you can do it yourself. I have answered the important parts only :)

Answer 2

To begin with, you can't store 10 numbers in a 3x3 matrix. Always make sure your specification makes sense before you start programming.

Taking the input is trivial, just use a nested loop to control where the read input ends up in your matrix. Example with a fixed array size:

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

#define x 3
#define y 3

int main (void)
{
  int arr[x][y];

  printf("Enter %d numbers: ", x*y);

  for(size_t i=0; i<x; i++)
  {
    for(size_t j=0; j<y; j++)
    {
      scanf(" %d", &arr[i][j]);
    }
  }

  for(size_t i=0; i<x; i++)
  {
    for(size_t j=0; j<y; j++)
    {
      printf("%d ", arr[i][j]);
    }
    printf("\n");
  }

  return 0;
}

(Note that this leaves a whole lot of junk like line feed characters behind in stdin. There's also no buffer overflow protection. See How to read / parse input in C? The FAQ for examples of how to read input properly.)

---

If you need a variable size matrix, you can use variable length arrays (VLA):

size_t x = 3; // some run-time value
size_t y = 3; // some run-time value
int arr[x][y];
... // then same code as above

Alternatively, you can use a dynamically allocated 2D array:

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

int main (void)
{
  size_t x = 3;
  size_t y = 3;
  int (*arr)[y] = malloc( sizeof(int[x][y]) );

  printf("Enter %zu numbers: ", x*y);

  for(size_t i=0; i<x; i++)
  {
    for(size_t j=0; j<y; j++)
    {
      scanf(" %d", &arr[i][j]);
    }
  }

  for(size_t i=0; i<x; i++)
  {
    for(size_t j=0; j<y; j++)
    {
      printf("%d ", arr[i][j]);
    }
    printf("\n");
  }

  free(arr);

  return 0;
}

Alternatively, if your compiler is from the Jurassic period, you can use old style "mangled" 2D arrays:

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

int main (void)
{
  size_t x = 3;
  size_t y = 3;
  int* mangled = malloc(x * y * sizeof *mangled);

  printf("Enter %zu numbers: ", x*y);

  for(size_t i=0; i<x; i++)
  {
    for(size_t j=0; j<y; j++)
    {
      scanf(" %d", &mangled[i*x + j]);
    }
  }

  for(size_t i=0; i<x; i++)
  {
    for(size_t j=0; j<y; j++)
    {
      printf("%d ", mangled[i*x + j]);
    }
    printf("\n");
  }

  free(mangled);

  return 0;
}

The above 4 alternatives are the only alternatives. You should not use "pointer-to-pointer look-up tables", there is absolutely no need for them here. Unfortunately, lots of bad books and bad teachers spread that technique. See Correctly allocating multi-dimensional arrays.

Answer 3

Try this:

int** InitMatrix()
{
    int** matrix = NULL;

    int n;
    if (scanf("%d", &n) > 0 && n > 0)
    {
        matrix = malloc(n * sizeof(int*));
        for (int i = 0; i < n; i++)
        {
            matrix[i] = malloc(n * sizeof(int));
            for (int j = 0; j < n; j++)
            {
                if (scanf("%d", &matrix[i][j]) == 0)
                {
                    // user input error; decide what to do here
                }
            }
        }
    }
    else
    {
        // user input error; decide what to do here
    }

    return matrix;
}