SC2055

if [[ $1 != foo || $1 != bar ]]
then 
  echo "$1 is not foo or bar"
fi

if [[ $1 != foo && $1 != bar ]]
then 
  echo "$1 is not foo or bar"
fi    

This is not a bash issue, but a simple, common logical mistake applicable to all languages.

[[ $1 != foo || $1 != bar ]] is always true:

• If $1 = foo then $1 != bar is true, so the statement is true.
• If $1 = bar then $1 != foo is true, so the statement is true.
• If $1 = cow then $1 != foo is true, so the statement is true.

[[ $1 != foo && $1 != bar ]] matches when $1 is not foo and not bar:

• If $1 = foo, then $1 != foo is false, so the statement is false.
• If $1 = bar, then $1 != bar is false, so the statement is false.
• If $1 = cow, then both $1 != foo and $1 != bar is true, so the statement is true.

This statement is identical to ! [[ $1 = foo || $1 = bar ]], which also works correctly.

None.