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?
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
ifcommand shall be the exit status of thethenorelsecompound-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.
if statement is zero, if none of the then/else branches were taken. if false; then false; fi leaves the exit status to 0, but if true; then false; fi leaves the exit status to 1. (even though in both cases the last command to run was false, and yes, I think it could return some other nonzero status too.)
is_match should probably just be is_match() { grep -qFe "$1" file.txt; } or is_match() { grep -qF -- "$1" file.txt; }, unless you do also want to store the matching line in $m (but then that would be poorly chosen name for that global reply variable)
$m a local variable.