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ath1410 wrote on Fri, Nov 22, 2002 at 05:04:55PM +0900 :
> 
> I checked the sites you refered, but did not still quite figure
> out (partly because of my poor English?) the way bash works the
> way it does. To make things simple, just take up one case here.
> $PS1=\\\\u           -4  returns \u
> the 1st \ escapes the 2nd \, and the 3rd one to the 4th one
> likewise.
> So  $PS1=\\\\u is something like $PS1='\\u' ( which actually
> returns \u, not user name.

Yes.

The following are identical:
PS=\\\\u
PS="\\\\u"


This is NOT the same:
PS='\\\\u'

Nor seemingly is this because of the single quoting which prevents
parameter expansion.  I honestly thought it _would_ be the same, but if
you say it doesn't work, then it doesn't work:
PS='\\u'

The trick is that bash parses each commandline twice (repeating someone
else's authoritative post from memory, so anybody more knowledgable,
please feel free to correct errors).  The first time it converts "\\" to
"\" and "\\" to "\", leaving you with "\\u".  The second time it
converts "\\" to "\", leaving you with "\u".  That is what then gets
used.

Blue skies...                   Todd
- -- 
           MandrakeSoft USA   http://www.mandrakesoft.com
          cat /boot/vmlinuz > /dev/dsp  #for great justice
   Cooker Version mandrake-release-9.1-0.1mdk Kernel 2.4.20-0.2mdk
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