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From: "Bart Schaefer" <schaefer@candle.brasslantern.com>
Message-Id: <991122201453.ZM23495@candle.brasslantern.com>
Date: Mon, 22 Nov 1999 20:14:53 +0000
In-Reply-To: <19991122190225.A8613@psidev.net>
Comments: In reply to Phil Pennock <phil@psidev.net>
        "expansion of nested parameters" (Nov 22,  7:02pm)
References: <19991122190225.A8613@psidev.net>
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To: Phil Pennock <phil@psidev.net>, Zsh Users <zsh-users@sunsite.auc.dk>
Subject: Re: expansion of nested parameters
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On Nov 22,  7:02pm, Phil Pennock wrote:
} Subject: expansion of nested parameters
}
} This is with zsh-3.0.7:
} -----------------------------< cut here >-------------------------------
} #!/bin/zsh -f
} a=foo
} wib_foo_ble=Wow
} c=\$wib_${a}_ble
} print ${(e)c}
} print ${(e)\$wib_${a}_ble}
} -----------------------------< cut here >-------------------------------
} 
} 1: Why does the second print statement print out the current process ID
}    from $$ instead of printing the same as the first print statement?

${(e)c} means:  Expand ${c}, then apply command substitution etc. to the
result of that expansion.

Hence ${(e)\$wib_${a}_ble} means: Expand ${\$wib_${a}_ble}, then ...

You'll note that ${\$} is the same as ${$} because the command-line parser
strips the backslash before the parameter-name interpreter ever gets around
to looking at it.  ${$} is the same as $$; the rest of the "wib_${a}_ble"
stuff is simply ignored once zsh recognizes $$.

} 2: Is there a way to do this without using an intermediate variable, as
}    I've done with $c and without using eval statements?

Yeah; it's in the FAQ, question 3.22.  What you're asking is a variation:

    print ${(e)a:+\$wib_${a}_ble}

should do it.

-- 
Bart Schaefer                                 Brass Lantern Enterprises
http://www.well.com/user/barts              http://www.brasslantern.com