Bash IF statement is not behaving as expected

Question

I cannot understand the logic of what's going on with this simple Bash script:

#!/bin/bash

# Check if file is present
which ./vendor/bin/non_existent_file &> /dev/null

printf "Exited with $?\n\n"

if [[ "$?" -eq 1 ]]; then
  echo "Number is One"
else
  echo "Number is not one"
fi

When the file is missing (non existent), the output is this:

Exited with 1

Number is not one

When the file is present, the output is this:

Exited with 0

Number is not one

???

Things I've tried:

if [ $? == 1 ]
if [ "$?" == 1 ]
if [[ "$?" == 1 ]]
if [[ $? -eq 1 ]]
if [[ "$?" = "1" ]]
if [[ "$?" == "1" ]]

Why is the IF statement always failing?

Somewhat related: [How to keep last exit status after test](https://unix.stackexchange.com/questions/209419/how-to-keep-last-exit-status-after-test) — steeldriver, Nov 28 '22 at 14:58
[Why not use "which"? What to use then?](https://unix.stackexchange.com/questions/85249/why-not-use-which-what-to-use-then) — glenn jackman, Nov 28 '22 at 15:19

score 3 · Accepted Answer · answered Nov 28 '22 at 15:43

which ./vendor/bin/non_existent_file &> /dev/null

This will run which and set $? to its exit status. (I'll assume for now that which works for you.)

printf "Exited with $?\n\n"

This will run printf, and set $? to its exit status.

if [[ "$?" -eq 1 ]]; then

So what gets tested is the exit status of printf.

You need to save the exit status to a temporary variable to avoid that, e.g.

which ./vendor/bin/non_existent_file &> /dev/null
ret=$?
printf 'which exited with status %d\n' "$ret"
if [[ $ret -ne 0 ]]; then
    printf "that was a falsy exit status"
fi

Though which as I know would search PATH for the named executable, but if you have a fixed path you're looking at, you could probably use [[ -x ./path ]] directly to see if the file there is executable. If you're looking for a program in PATH, you may want to see Why not use "which"? What to use then? for caveats and corner cases.

score 1 · Answer 2 · answered Nov 28 '22 at 15:40

Thanks to the comments by @Kusalananda and @SottoVoce it becomes clear:

$? stores the result of the previous command, which includes printf.

So running printf "Exited with $?\n\n" changes $? to 0 by virtue of it having executed successfully.

It would be better to store the result of the execution in another variable to prevent this sort of confusion from happening.

Bash IF statement is not behaving as expected

2 Answers2