A junior administrator is experimenting with arithmetic loops in a Bash script. Examine the fragment below:
#!/usr/bin/env bash
num=2
until (( num++ >= 5 )); do
echo -n "$num "
done
echo # prints a final newline
The script is run under Bash 5.x with default settings and exits normally. What exact sequence is written to standard output before the shell prompt returns?
3 4 5
2 3 4 5 6
3 4 5 6
The loop never terminates because the condition is written incorrectly
An until loop keeps executing its body while the test command returns a non-zero exit status and stops as soon as the test succeeds (exit status 0). The test here is the arithmetic expression num++ >= 5, which is evaluated before each iteration. Because the postfix increment operator returns the variable's current value and then increments it, the first comparison checks 2 >= 5, which is false (result 0 → exit status 1). The loop therefore enters its body and prints the now-incremented value 3. The next two tests evaluate 3 >= 5 and 4 >= 5, both still false, so the body executes and prints 4 and then 5. When num reaches 5, the following test compares 5 >= 5; the expression evaluates to 1, giving an exit status 0, so the loop terminates and the body is not executed again. Consequently, only the numbers 3 4 5 are produced, followed by the newline from the final echo. The other answer choices either include the pre-increment value 2, continue one step beyond the terminating condition, or ignore that the loop stops once the test succeeds.
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What is the difference between 'until' and 'while' loops in Bash?
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How does the postfix increment operator 'num++' work in Bash arithmetic?
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Why does the script output start with 3 instead of 2?
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