[SOLVED] Why is [[ ! ! expr ]] equivalent to [[ ! expr ]] in bash?

Issue

In bash, I don't understand why the 3rd command isn't true:

[[ 1 -eq 1 ]]         # $? is 0
[[ ! 1 -eq 1 ]]       # $? is 1
[[ ! ! 1 -eq 1 ]]     # $? is 1 (??)
[[ ! ( ! 1 -eq 1 ) ]] # $? is 0

It seems to do the same thing if I replace 1 -eq 1 with any true expression, and negate with any false expression.

Solution

When parsing !, bash sets a flag to the following expression to tell it to invert its result. Because of this, parsing two or more ! is equivalent to parsing a single !. I proposed a patch to fix this, and it got accepted.

Answered By - vimene