Re: Is this a bug in [[ -f ]]?

[suvayu ali]

Hello Eric,

On Fri, Aug 19, 2011 at 5:04 PM, Eric Blake <eblake@redhat.com> wrote:
> On 08/19/2011 08:45 AM, Suvayu Ali wrote:
>>
>> I am trying to test if a file exists and then source it. My problem is
>> the test succeeds even if the variable is empty! If I pass no argument
>> at all, it still succeeds. To give you an example:
>>
>> $ unset bla
>> $ [ -f $bla ]&&  echo yes
>> yes
>> $ [ -f  ]&&  echo yes
>> yes
>
> Both expected behaviors, and evidence of your lack of quoting.
>
> Remember, the behavior of [] depends on how many arguments are present.
>
> [ -f "$bla" ] (note the "") - guarantees that there are exactly two
> arguments, so it proceeds with the two-argument test where -f is the
> operator and "$bla" is the file name.
>
> [ -f ] (which is the same as [ -f $bla ] if $bla is empty, note the lack of
> "") - there is exactly one argument, so it proceeds with the one-argument
> test of whether the argument (the literal string -f) is empty (it is not).
>
> Furthermore, [ -f $bla ] is different than [[ -f $bla ]].  [ is a POSIX
> utility, and mandated to do all argument word expansion before [ ever gets a
> chance to see what arguments it was given - if $bla is empty or has spaces,
> you changed the number of arguments that are given to [.  [[ is a bash (and
> ksh) extension that is part of the shell syntax (similar to how () for
> subshells is part of the syntax), thus it knows how many words,
> _pre-expansion_, were present, and the fact that $bla was unquoted is not a
> problem, [[ -f $bla ]] is a safe way to check if $bla is a file even if $bla
> is empty or contains spaces.
>

Thanks a lot for the very clear explanation. I'll be careful in the future. :)

-- 
Suvayu

Open source is the future. It sets us free.