On 10/10/10 8:46 PM, Devin Teske wrote:
On Oct 10, 2010, at 4:51 PM, Garance A Drosihn <dro...@rpi.edu
<mailto:dro...@rpi.edu>> wrote:
The latter does not cause an error. Try it:
# [ "-n" = x ] ; echo $?
1
# [ -e = "no" ] ; echo $?
1
# [ -e = -n ] ; echo $?
1
1 is error. 0 is success.
--
Um, yes, true. I know that. What I'm saying is that
the command works as you'd want it to work. It does
not hit a parsing error. The double-quotes did not
change how the command behaved. You deleted the
context of what I was replying to when I said the
above. Looking at the examples I gave there, it probably
would have been clearer if I had typed the exact same
command with and without the double-quotes. Eg:
On 10/10/10 7:09 PM, Devin Teske wrote:
> However, enclosing the argument (as the 'x$foo'
> portion is really just the first argument to the
> '[' built-in) in quotes:
>
> [ "$foo" = x ]
>
> makes it so that the expansion is taken as:
>
> [ "-n" = x ]
>
> rather than:
>
> [ -n = x ]
>
> The former not causing an error, while the latter does.
Your second example does not cause an error. Try it:
# [ "-n" = "-n" ] ; echo $?
0
# [ "-n" = x ] ; echo $?
1
Compared to the double-quote-less:
# [ -n = "-n" ] ; echo $?
0
# [ -n = x ] ; echo $?
1
--
Garance
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