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Bug or misundertanding from my part?
- X-seq: zsh-workers 21733
- From: DervishD <zsh@xxxxxxxxxxxx>
- To: Zsh Workers <zsh-workers@xxxxxxxxxx>
- Subject: Bug or misundertanding from my part?
- Date: Mon, 19 Sep 2005 16:32:35 +0200
- Mail-followup-to: Zsh Workers <zsh-workers@xxxxxxxxxx>
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- Organization: DervishD
Hi all :)
In the ZSH manual for 4.2.5 we can read:
{ TRY-LIST } always { ALWAYS-LIST }
First execute TRY-LIST. Regardless of errors, or break, continue,
or return commands encountered within TRY-LIST, execute
ALWAYS-LIST. Execution then continues from the result of the
execution of TRY-LIST; in other words, any error, or break,
continue, or return command is treated in the normal way, as if
ALWAYS-LIST were not present. The two chunks of code are referred
to as the `try block' and the `always block'.
So, this code should print "try list" and "always list", but it
doesn't:
#!/bin/zsh
emulate -L zsh
{
print "try list"
return 13
} always {
print "always list"
return 0
}
print "What?"
The code above just prints "try list" and exits with a return
code of "13", but the manual says that the ALWAYS-LIST is executed
"Regardless of [...] return commands encountered [...]". Am I doing
something wrong? I'm a bit puzzled because the above works if I
change 'return 13' for 'false'.
Thanks a lot in advance :)
Raúl Núñez de Arenas Coronado
--
Linux Registered User 88736 | http://www.dervishd.net
http://www.pleyades.net & http://www.gotesdelluna.net
It's my PC and I'll cry if I want to...
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