SC2252

You probably wanted && here, otherwise it's always true.

Problematic code:

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

Correct code:

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

Rationale:

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

[ "$1" != foo ] || [ "$1" != bar ] is always true (when foo != bar):

• 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 neither foo nor 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 (by De Morgan's law)

This warning is equivalent to SC2055 and SC2056, which trigger for intra-test expressions and arithmetic contexts respectively.

Exceptions

Rare.