How can I print multidimensional arrays in C?

Question

#include <stdio.h>

int main(){

  int number[3][4] = {
    {10,20,30,40},
    {15,25,35,45},
    {47,48,49,50},
  };
      printf("%s",number[]);
  return 0;
}

Ok, here's my code and what I wonder is how can I print out those arrays that I created. Do I have to use for loop or is there another way?

P.S: Im a newbie.

You have to write this yourself using a loop, just like with one-dimensional arrays. — Thomas, Feb 14 '20 at 17:24
Best way to do it is to have 2 for-loops, basically it would be one for columns and another for rows. — Vrej, Feb 14 '20 at 17:31

Vlad from Moscow · Accepted Answer · 2020-02-14T17:53:51.837

Use nested for loops. For example

for ( size_t i = 0; i < 3; i++ )
{
    for ( size_t j = 0; j < 4; j++ )
    {
        printf( "%2d ", number[i][j] );
    }
    putchar( '\n' );
}

Pay attention to that it is a bad idea to use magic numbers like 3 and 4. Instead use named constants.

As for the conversion specifier %s then it is used to output one-dimensional character arrays that contain strings

Here is a demonstrative program

#include <stdio.h>

int main(void) 
{
    enum { M = 3, N = 4 };
    int number[M][N] =
    {
        { 10, 20, 30, 40 },
        { 15, 25, 35, 45 },
        { 47, 48, 49, 50 },
    };

    for ( size_t i = 0; i < M; i++ )
    {
        for ( size_t j = 0; j < N; j++ )
        {
            printf( "%2d ", number[i][j] );
        }
        putchar( '\n' );
    }

    return 0;
}

Its output is

10 20 30 40 
15 25 35 45 
47 48 49 50

To output a two-dimensional integer array as a one-dimensional integer array apply casting to the array designator to the type int * or const int *.

For example

#include <stdio.h>

int main(void) 
{
    enum { M = 3, N = 4 };
    int number[M][N] =
    {
        { 10, 20, 30, 40 },
        { 15, 25, 35, 45 },
        { 47, 48, 49, 50 },
    };

    const int *p = ( const int * )number;

    for ( size_t i = 0; i < M * N; i++ )
    {
        printf( "%2d ", p[i] );
    }
    putchar( '\n' );

    return 0;
}

The program output is

10 20 30 40 15 25 35 45 47 48 49 50

And vice versa to print a one dimensional array as a two-dimensional array use the following approach.

#include <stdio.h>

int main(void) 
{
    enum { N = 12 };
    int number[N] = { 10, 20, 30, 40, 15, 25, 35, 45, 47, 48, 49, 50 };

    size_t rows = 3, cols = 4;

    for ( size_t i = 0; i < rows; i++ )
    {
        for ( size_t j = 0; j < cols; j++ )
        {
            printf( "%2d ", number[i * cols + j] );
        }
        putchar( '\n' );
    }           

    return 0;
}

The program output is

10 20 30 40 
15 25 35 45 
47 48 49 50

score 1 · Answer 2 · answered Feb 14 '20 at 17:28

%s expects a null terminated string of chars. You print arrays by iterating through them:

for (i = 0; i < 3; i++)
{
    for (i2 = 0; i2 < 4; i2++)
    {
        printf("%d ", number[i][i2]);
    }

    putchar('\n');
}

It helps to do this in row-major order if performance matters.

score 1 · Answer 3 · answered Feb 14 '20 at 17:32

So there are several problems with your code.

You're missing an array index in printf("%s",number[]);
You're using the wrong format specifier. It should be %d, not%s. The %d doesn't just act as a placeholder; it also tells C how to properly convert your number into a displayable form.

To print the array, you need a nested for loop. Your program should look something like this:

int main(){

  int number[3][4] = {
    {10,20,30,40},
    {15,25,35,45},
    {47,48,49,50},
  };

  for (int i=0; i<3; i++) {
    for (int j=0; j<4; j++) {
      printf("%d ",number[i][j]);
    }
    printf("\n");
  }

  return 0;
}

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How can I print multidimensional arrays in C?

3 Answers3