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Home/ Questions/Q 7911719
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T13:21:17+00:00 2026-06-03T13:21:17+00:00

I want to output a string by adding random integer to a variable to

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I want to output a string by adding random integer to a variable to create the string. Bash however, just adds the numbers together.

#!/bin/bash
b=""
for ((x=1; x<=3; x++))
do
 number=$RANDOM
 let number%=9
 let b+=$number
done
echo ${b}

Say every random number is 1, the script will output 3 instead of 111.
How do I achieve the desired result of 111?

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  1. Editorial Team
    Editorial Team
    2026-06-03T13:21:18+00:00Added an answer on June 3, 2026 at 1:21 pm

    There are several possibilities to achieve your desired behavior. Let’s first examine what you’ve done:

    let b+=$number
    

    Running help let:

    let: let ARGUMENT...
        Evaluate arithmetic expressions.
    

    That explains why let b+=$number performs an integer addition (1, 2, 3) of $number to b instead of string concatenation.

    Simply remove let and the desired behavior 1, 11, 111 will occur.

    The other method to perform string concatenation:

    b="$b$number"
    

    Yes, simply “let b become the result of concatenating b and number.

    As a side note, b="" is equivalent to b= as "" is expanded to an empty string. Module operation on a variable can be done with arithmetic expansion: number=$((RANDOM%9)).

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