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Re: escaping


From: Chet Ramey
Subject: Re: escaping
Date: Wed, 3 Oct 2001 13:38:05 -0400

> simple problem, but a basic problem nonetheless
> 
> consider
> 
> echo \z #backslash is a special character to the shell, so it is not printed
> result: z
> echo "\z" #when using double quotes, every special character is turned off
> except variable and argument substitution ($), backticks and backslash.
> why does the last statement output: \z instead of (what i expected) the same
> as the former command (a single z)???

You should have read the manual.  It says, in the QUOTING section:

       The backslash
       retains its special meaning only when followed by  one  of
       the  following  characters:  $,  `, ", \, or <newline>

-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer
( ``Discere est Dolere'' -- chet)

Chet Ramey, CWRU    address@hidden    http://cnswww.cns.cwru.edu/~chet/



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