Displaying bash array

Question

#!/bin/bash
declare -a array
printf 'Matrix size:' ;
read n;
printf '\n';
      for ((i=1;i<=n;i++))
        do
      for((j=1;j<=n;j++))
        do
          printf 'x[%d][%d]=' ${array[i][j]};
          read array[i][j];
     done
     done
     echo "Initial matrix:"
        for((i=1;i<=$n;i++))
          do 
        for ((j=1;j<=n;j++))
          do
       printf '%d' ${array[i][j]};
           printf '\n';
        done
        done

A can't display bash array and don't understand where is my mistake. For example I have: n=3....I enter numbers in array (1-9) When I displaying: Initial matrix: 3 3 3 6 6 6 9 9 9 Thx

Answer 1

bash doesn't have multidimensional arrays; when you reference array[i][j] , the [j] is ignored. bash also has other limitations that tend to make it unsuitable for this sort of thing, such as not supporting floating point math (natively, anyway).

If you need to fake a multidimensional array in bash, you can fake it by using array[i*n+j] to store array[i][j]:

#!/bin/bash
declare -a array
read -p 'Matrix size: ' n
for ((i=1; i<=n; i++)); do
    for ((j=1; j<=n; j++)); do
        read -p "x[$i][$j]=" array[i*n+j]
    done
done

echo "Initial matrix:"
for ((i=1; i<=n; i++)); do 
    for ((j=1; j<=n; j++)); do
        printf '%d ' ${array[i*n+j]}
    done
    printf '\n'
done

Note that I've done some additional cleanup on your code:

• Don't put semicolons at the end of a line, they're redundant in a shell script.
• Print prompts with read -p instead of printf .
• The input loop was printing array[i][j] (ie the array's contents) where it should've printed i and j.
• You don't need to use printf to interpolate variables in a string, just embed $i in a double-quoted string (or use ${i} to avoid ambiguity about where the end of the variable name is).
• Finally, when printing the array I made it put a newline after each row, rather than after each element.

Answer 2

It's possible to use bash 4's associative arrays to get multidimensional arrays... sorta:

#!/bin/bash

declare -A array

printf "Matrix size: "
read -r n
for ((i = 0; i < n; i++)); do
    for ((j = 0; j < n; j++)); do
        printf "array[$i][$j] = "
        read -r val
        array["${i}_${j}"]=$val
    done
done

echo "Initial matrix:"
for key in "${!array[@]}"; do
    val=${array[$key]}
    echo "$key $val"
done

However, there is no true multidimensional array support available in bash.

Unlike the answer using standard numerically-indexed arrays, this approach doesn't require the dimensions of the array to be known before it can be read.

Answer 3

All of Gordon's remarks are absolutely correct, for the sake of completeness I'll just add that ksh93 (I am not certain about ksh88 ) does support multidimensional arrays (but it's a notoriously un(der)documented feature), so you can do this natively with ksh :

matrix.sh :

#!/bin/ksh

printf "Matrix size: "
read n
for ((i = 0; i < n; i++)); do
        for ((j = 0; j < n; j++)); do
                printf "array[$i][$j] = "
                read array[i][j]
        done
done

echo "Initial matrix:"
for ((i = 0; i < ${#array[@]}; i++)); do
        for k in "${array[i][@]}"; do
                printf "${k} "
        done
        echo
done

Example:

$ ./matrix.sh
Matrix size: 2
array[0][0] = 3
array[0][1] = 4
array[1][0] = 5
array[1][1] = 6
Initial matrix:
3 4
5 6