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Why is the bash builtin `test' slightly different from /bin/test?

The specific example used here is [ ! x -o x ], which is false.



Bash's builtin `test' implements the Posix.2 spec, which can be

summarized as follows (the wording is due to David Korn):

   

Here is the set of rules for processing test arguments.

  

    0 Args: False

    1 Arg:  True iff argument is not null.

    2 Args: If first arg is !, True iff second argument is null.

	    If first argument is unary, then true if unary test is true

	    Otherwise error.

    3 Args: If second argument is a binary operator, do binary test of $1 $3

	    If first argument is !, negate two argument test of $2 $3

	    If first argument is `(' and third argument is `)', do the

	    one-argument test of the second argument.

	    Otherwise error.

    4 Args: If first argument is !, negate three argument test of $2 $3 $4.

	    Otherwise unspecified

    5 or more Args: unspecified.  (Historical shells would use their

    current algorithm).

   

The operators -a and -o are considered binary operators for the purpose

of the 3 Arg case.

   

As you can see, the test becomes (not (x or x)), which is false.