Return value of function in UNIX

Question

I do not understand the following expression.

function abc(){

..............

...............

[[ -f $filename]] && return 0 || return 1

}

As per tutorial if there is a file exists with filename variable name then this function returns 1 otherwise it returns 0.

I understand && || operator ,but how is this statement getting the desire result?

As per me,In case [[ -f $filename ]] evaluates false ,then one statement of && is false then result of and is false.Now it goes to OR and if first operand is 0 it returns result of second operand so it should return 1,but instead it is returning 0.

How is this being evaluated?

Answer 1

Both return statements on that last line of the function can be removed.

[[ -f "$filename" ]]

This is the last statement in the function with both return's removed (note the quoted variable expansion and the added space before ]]). The "exit value" of the function will be the result of this statement.

If the file $filename exists, the function will exit with a value of zero (signifying "success", "yes", "ok" etc.), otherwise it will exit with a value of one (or more generally, non-zero, signifying "failure" of some kind).

---

Don't mix || and && on the same line unless you know what it does. In a command line as

command1 && command2 || command3

the last command would be executed if either of the previous commands failed (returned non-zero).

It's better to write

if command1; then
    command2
else
    command3
fi

if this is what you meant.

This matters in commands like

[[ -f "$filename" ]] && echo "exists" || touch "$filename"

This would try to execute the touch command if the echo failed, which it may do if there's nowhere to output the string to (a write error occurs).