Bash error in quote handling
Arthur I Schwarz
Tue Feb 15 13:56:00 GMT 2005
Thank your for your explanation.
Pechtchanski To: Arthur I Schwarz <Arthur_I_Schwarz@raytheon.com>
<firstname.lastname@example.org cc: email@example.com
.edu> Subject: Re: Bash error in quote handling
On Mon, 14 Feb 2005, Arthur I Schwarz wrote:
> I think this is an error?
Huh? What is? What exactly are the commands you're trying to run? What
exactly is the output? What is wrong with the output?
Let me try to guess (below):
> a. star="*"; echo $star
> b. star=*; echo $star
You're saying that the above commands produce the contents of the
directory, and that this is correct.
> c. star='*'; echo $star
> d. star=\*; echo $star
The above commands produce the contents of the directory, and you think
this is incorrect (should be a literal "*").
> e. star='\*'; echo $star
> f. star="\*"; echo $star
The above commands produce the string "\*", and you think it should be
AFAIK, the above is all expected behavior.
The first four commands (a.-d.) set the variable "star" to the same exact
value, namely, the literal "*". When the shell sees $star, it expands the
contents of it (which is a "*" wildcard), so you get the contents of the
directory. The manner of quoting you used when assigning the value of the
variable is irrelevant.
If you want echo to produce a literal "*", use 'echo "$star"' (double
quotes to allow variable substitution but prevent shell expansion).
Frankly, at the moment I can't think of a reason why the last two commands
produce "\*" (it's not a nullglob issue, as I first thought). I'm
reasonably sure it's also expected, but I'll investigate the actual reason
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