How to evaluate a boolean variable in an if block in bash?

I have defined the following variable:

myVar=true

now I'd like to run something along the lines of this:

if [ myVar ]
then
    echo "true"
else
    echo "false"
fi

The above code does work, but if I try to set

myVar=false

it will still output true. What might be the problem?

edit: I know I can do something of the form

if [ "$myVar" = "true" ]; then ...

but it is kinda awkward.

Thanks

bash doesn't know boolean variables, nor does test (which is what gets called when you use [).

A solution would be:

if $myVar ; then ... ; fi

because true and false are commands that return 0 or 1 respectively which is what if expects.

Note that the values are "swapped". The command after if must return 0 on success while 0 means "false" in most programming languages.

SECURITY WARNING: This works because BASH expands the variable, then tries to execute the result as a command! Make sure the variable can't contain malicious code like rm -rf /

Note that the if $myVar; then ... ;fi construct has a security problem you might want to avoid with

case $myvar in
  (true)    echo "is true";;
  (false)   echo "is false";;
  (rm -rf*) echo "I just dodged a bullet";;
esac

You might also want to rethink why if [ "$myvar" = "true" ] appears awkward to you. It's a shell string comparison that beats possibly forking a process just to obtain an exit status. A fork is a heavy and expensive operation, while a string comparison is dead cheap. Think a few CPU cycles versus several thousand. My case solution is also handled without forks.