Why does $((true == false)) evaluate to 1 in bash?

Question

Why does bash have the following behavior?

echo $((true == false))
1

I would have thought that this wou

Answer 1

OK, so, it seems there is a lot of confusion about what $((...)) really does.

It does an arithmetic evaluation of its operands, barewords are variables, not commands or string literals (ie, true is really $true), and anything which is not a number is 0. The == operator compares two numbers, and returns 1 if they are equal.

Which is why $((true == false)) is 1: there is no true or false environment variables in the environment, which means both $true and $false evaluate to the empty string, therefore 0 in arithmetic context!

To be complete, you can also use command substitution in arithmetic context... For instance:

$ echo $((`echo 2`))
2
$ echo $((3 + $(echo 4)))
7
$ a=3
$ echo $((a + $(echo 4)))
7
# undefine a
$ a=
$ echo $((a + $(echo 4)))
4
$ a="Hello world"
$ echo $((a + 1))
1
# undefine a again
$ a=
$ echo $(($(echo $a)))
0