Re: bash incompatible with sh,maybe this is a bug

[Andreas Schwab]

showrun.lee <showrun.lee@gmail.com> writes:

> [  !  "$test"  -a "$test1"  ] && test="papa";

This is a test with four arguments, thus the rules are:

  • If $1 is ’!’, negate the three-argument test of $2, $3, and $4.
  • If $1 is ’(’ and $4 is ’)’, perform the two-argument test of $2 and $3.
  • Otherwise, the results are unspecified.

Since "$test" -a "$test1" is false, the test comes out as true.

Using test with a combination of more than one primary operator is not
recommended.

> I think ,this is a bug in bash.and i test it on many freebsd release 4.11,
> 6.1,6.2 with default bash version,this is same.

Your /bin/sh is not POSIX compliant.

Andreas.

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Andreas Schwab, SuSE Labs, schwab@suse.de
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