print equally spaced elements of a 2D array with printf - c

Question

I am trying to print a 2d array that has a maximum of 3 digit numbers that are aligned when printed. For example, with a simple printf, it looks like this:
[0, 232, 20, 96, 176, 0, 0]
[0, 0, 24, 0, 0, 176, 0]
[0, 0, 0, 0, 0, 0, 0]
I would like it to be printed with all the commas aligned along the columns with additional whitespace, like this:
[ 0, 232, 20, 96, 176, 0, 0]
[ 0, 0, 24, 0, 0, 176, 0]
[ 0, 0, 0, 0, 0, 0, 0]
How can I do this with printf?

Answer 1

You can use the width prefix to specify the minimum width for the printf conversions: printf("%4d", x); will print int variable x padded to the left with enough spaces to produce at least 4 characters.
If you know the maximum width of any number in the array, you can hardcode this number in the format string. Otherwise you can compute the required width and use %*d and pass an extra argument to specifying the computed width.
Here is a modified version:
#include <stdio.h>
int main(void) {
#define M 3
#define N 7
int a[M][N] = {
{ 0, 232, 20, 96, 176, 0, 0 },
{ 0, 0, 24, 0, 0, 176, 0 },
{ 0, 0, 0, 0, 0, 0, 0 },
};
int width = 0;
/* compute the required width */
for (size_t i = 0; i < M; i++) {
for (size_t j = 0; j < N; j++) {
int w = snprintf(NULL, 0, "%d", a[i][j]);
if (width < w) {
width = w;
}
}
}
/* print the arrays */
for (size_t i = 0; i < M; i++) {
printf("[");
for (size_t j = 0; j < N; j++) {
if (j != 0) printf(", ");
printf("%*d", width, a[i][j]);
}
printf("]\n");
}
return 0;
}
Output:
[ 0, 232, 20, 96, 176, 0, 0]
[ 0, 0, 24, 0, 0, 176, 0]
[ 0, 0, 0, 0, 0, 0, 0]

Answer 2

Here you are.
#include <stdio.h>
int main(void)
{
enum { M = 3, N = 7 };
int a[M][N] =
{
{ 0, 232, 20, 96, 176, 0, 0 },
{ 0, 0, 24, 0, 0, 176, 0 },
{ 0, 0, 0, 0, 0, 0, 0 }
};
int width = 4;
for ( size_t i = 0; i < M; i++ )
{
putchar( '[' );
for ( size_t j = 0; j < N; j++ )
{
if ( j != 0 ) putchar( ',' );
printf( "%*d", width, a[i][j] );
}
printf( "]\n" );
}
putchar( '\n' );
return 0;
}
The program output is
[ 0, 232, 20, 96, 176, 0, 0]
[ 0, 0, 24, 0, 0, 176, 0]
[ 0, 0, 0, 0, 0, 0, 0]
You can change the value of the variable width if you want for example to output larger values than values consisting from 3 digits. That is the output is enough flexible.
If you want to place the array output in a separate function then the corresponding function can look the following way provided that the compiler supports variable length arrays.
#include <stdio.h>
void format_output( size_t m, size_t n, int a[m][n], int width )
{
for ( size_t i = 0; i < m; i++ )
{
putchar( '[' );
for ( size_t j = 0; j < n; j++ )
{
if ( j != 0 ) putchar( ',' );
printf( "%*d", width, a[i][j] );
}
printf( "]\n" );
}
}
int main(void)
{
enum { M = 3, N = 7 };
int a[M][N] =
{
{ 0, 232, 20, 96, 176, 0, 0 },
{ 0, 0, 24, 0, 0, 176, 0 },
{ 0, 0, 0, 0, 0, 0, 0 }
};
int width = 4;
format_output( M, N, a, width );
putchar( '\n' );
return 0;
}
The program output is the same as shown above that is
[ 0, 232, 20, 96, 176, 0, 0]
[ 0, 0, 24, 0, 0, 176, 0]
[ 0, 0, 0, 0, 0, 0, 0]

Answer 3

use this format of printf printf("%4d", arr[i][j]);
int main()
{
int arr[3][7] = { {0, 232, 20, 96, 176, 0, 0}
,{0, 0, 24, 0, 0, 176, 0}
,{0, 0, 0, 0, 0, 0, 0} };
for (int i = 0; i < 3; i++)
{
printf("[");
for (int j = 0; j < 7; j++)
{
if (j < 6)
printf("%4d,", arr[i][j]);
if (j == 6)
printf("%4d", arr[i][j]);
}
printf("]");
printf("\n");
}
}
PS:amount of space can be changed as needed ,with changing "%4d".