OR logical operator in bash -

i have dumb problem can't wrap head around it

if [[ false || false ]] ;         echo 'true' else         echo 'false' fi 

as per http://tldp.org/ldp/abs/html/comparison-ops.html

-o logical or

exp1 -o exp2 returns true if either exp1 or exp2 true.

these similar bash comparison operators && , ||, used within double brackets. [[ condition1 && condition2 ]]

so if both false should return false? why prints 'true'?

you should run not part of conditional command '[[ ]]':

if false || false;         echo 'true' else         echo 'false' fi 

as testing falses within [[ , ]]:

if [[ ! 1 || ! 1 ]];         echo 'true' else         echo 'false' fi 

noting [[ false ]] equivalent [[ -n false ]] makes true condition.

if make more apparent , valid conditional test (( )) this:

if (( 0 || 0 ));         echo 'true' else         echo 'false' fi