bash "if [ false ];" returns true instead of false -- why?

Why does the following output True?

#!/bin/sh

if [ false ]; then
    echo "True"
else
    echo "False"
fi

This will always output True even though the condition would seem to indicate otherwise. If I remove the brackets [] then it works, but I do not understand why.

Solution 1:

You are running the [ (aka test) command with the argument "false", not running the command false. Since "false" is a non-empty string, the test command always succeeds. To actually run the command, drop the [ command.

if false; then
   echo "True"
else
   echo "False"
fi

Solution 2:

A Quick Boolean Primer for Bash

The if statement takes a command as an argument (as do &&, ||, etc.). The integer result code of the command is interpreted as a boolean (0/null=true, 1/else=false).

The test statement takes operators and operands as arguments and returns a result code in the same format as if. An alias of the test statement is [, which is often used with if to perform more complex comparisons.

The true and false statements do nothing and return a result code (0 and 1, respectively). So they can be used as boolean literals in Bash. But if you put the statements in a place where they're interpreted as strings, you'll run into issues. In your case:

if [ foo ]; then ... # "if the string 'foo' is non-empty, return true"
if foo; then ...     # "if the command foo succeeds, return true"

So:

if [ true  ] ; then echo "This text will always appear." ; fi;
if [ false ] ; then echo "This text will always appear." ; fi;
if true      ; then echo "This text will always appear." ; fi;
if false     ; then echo "This text will never appear."  ; fi;

This is similar to doing something like echo '$foo' vs. echo "$foo".

When using the test statement, the result depends on the operators used.

if [ "$foo" = "$bar" ]   # true if the string values of $foo and $bar are equal
if [ "$foo" -eq "$bar" ] # true if the integer values of $foo and $bar are equal
if [ -f "$foo" ]         # true if $foo is a file that exists (by path)
if [ "$foo" ]            # true if $foo evaluates to a non-empty string
if foo                   # true if foo, as a command/subroutine,
                         # evaluates to true/success (returns 0 or null)

In short, if you just want to test something as pass/fail (aka "true"/"false"), then pass a command to your if or && etc. statement, without brackets. For complex comparisons, use brackets with the proper operators.

^{_{And yes, I'm aware there's no such thing as a native boolean type in Bash, and that if and [ and true are technically "commands" and not "statements"; this is just a very basic, functional explanation.}}

Solution 3:

I found that I can do some basic logic by running something like:

A=true
B=true
if ($A && $B); then
    C=true
else
    C=false
fi
echo $C

Solution 4:

Using true/false removes some bracket clutter...

#! /bin/bash    
#  true_or_false.bash

[ "$(basename $0)" == "bash" ] && sourced=true || sourced=false

$sourced && echo "SOURCED"
$sourced || echo "CALLED"

# Just an alternate way:
! $sourced  &&  echo "CALLED " ||  echo "SOURCED"

$sourced && return || exit

Solution 5:

as noted by @tripleee, this is tangential, at best.

still, in case you arrived here searching for something like that (as i did), here is my solution

having to deal with user acessible configuration files, i use this function :

function isTrue() {
    if [[ "${@^^}" =~ ^(TRUE|OUI|Y|O$|ON$|[1-9]) ]]; then return 0;fi
    return 1
    }

wich can be used like that

if isTrue "$whatever"; then..

You can alter the "truth list" in the regexp, the one in this sample is french compatible and considers strings like "Yeah, yup, on,1, Oui,y,true to be "True".

note that the '^^' provides case insensivity