Difference between if [ ... and test ... statement in bash

Question

Consider the following:

echo "hello" > file.txt
is_match1 () {
  local m
  m=$(cat "file.txt" | grep -F "$1")
  if [ -z "$m" ]; then
    return 1
  fi
}
is_match2 () {
  local m
  m=$(cat "file.txt" | grep -F "$1")
  test -z "$m" && return 1
}
is_match1 "hello"
echo "$?"
0
is_match2 "hello"
echo "$?"
1

Why does is_match2 return 1?

Answer 1

Under the circumstances of your question, m gets the value hello in both functions.

Now look at

test -z "$m" && return 1

What should happen here? The -z test is false, right? So return 1 does not execute. Instead, what does the function return? Every function returns the value of $? at the end. In this case, that value is 1, the result of the && list.

What you might have wanted to test is

if [ -z "$m" ]; then return 1; fi

compared to

if test -z "$m"; then return 1; fi

The exit status of both these if statements are zero when $m is non-empty, since none of the statements' branches were taken.

From the POSIX standard:

The exit status of the if command shall be the exit status of the then or else compound-list that was executed, or zero, if none was executed.

---

Note that we may condense both your functions down into

is_match () {
    grep -q -F -e "$1" file.txt
}

Here, we let grep provide the exit status to the caller. We also do not necessarily read the file.txt file to the end, as grep -q quits as soon as it finds a match.