Why does double brackets notation escape strings in bash? -
given following example code:
var="foo" if [[ ${var} != 'bar' ]]; then...
if run piece of code in bash debug mode (bash -x
), reads out follows:
[[ foo != \b\a\r ]]
it still passes correctly, why show escapes? there better way this?
set -x
ignores particular method of escaping , quoting, , constructs own equivalent representation of strings. shouldn't care exact format.
'bar'
, \b\a\r
encodes exact same string, , that's matters.
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