Crucial Exams

CompTIA Linux+ XK0-006 (V8) Practice Question

During a code review you spot the following Bash snippet that is supposed to keep a running counter:

#!/usr/bin/env bash
count=0

increment() {
    local count=$((count + 1))
}

increment
echo "$count"

The developer expects the script to print 1, but it prints 0. Which single change will make the script print 1 while remaining portable to any POSIX-compliant /bin/sh and without starting a subshell?

  • Replace local with declare -g -i count=$((count + 1)) inside the function.

  • Delete the word local so the function assigns directly to the existing count variable.

  • Add export count before the function declaration to make the variable global.

  • Capture the function's output: count=$(increment) instead of calling the function directly.

CompTIA Linux+ XK0-006 (V8)
Automation, Orchestration, and Scripting
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