From: Bart Schaefer <schaefer@brasslantern.com>
To: zsh-users@zsh.org
Subject: Re: Could someone clarify how math functions work?
Date: Thu, 18 Dec 2014 21:34:50 -0800 [thread overview]
Message-ID: <141218213450.ZM28822@torch.brasslantern.com> (raw)
In-Reply-To: <54939F50.50102@gmx.com>
On Dec 18, 10:45pm, Eric Cook wrote:
} Subject: Could someone clarify how math functions work?
}
} zsh -c 'add() ( for arg; do (( n += arg )); done; print n: $n );
} functions -M add; print results: $(( add(1,2,3) ))'
}
} Outputs:
} n: 6
} results: 3
}
} where as:
} zsh -c 'add() { local arg n; for arg; do (( n += arg )); done; print n:
} $n }; functions -M add; print results: $(( add(1,2,3) ))'
}
} Outputs:
} n: 6
} results: 6
}
} Is that expected behavior? If so, could you explain why?
The following might illustrate:
zsh -c 'add() ( for arg; do (( n += arg )); done; print n: $n );
functions -M add; print results: $(( add(3,2,1) ))'
zsh -c 'add() ( for arg; do (( n += arg )); done; print n: $n );
functions -M add; print results: $(( add(3,1,2) ))'
When you define add() with parens ( ) around the function body, you
are running the function body in a subshell. The "last arithmetical
expression evaluated" IN THE CURRENT SHELL is the processing of the
argument list of the call to add(), which is done left-to-right and
is therefore "3" in the original example.
When you define add() with braces { } you are running the function
body in in the current shell, so the last expression is the last
assignment in the for-loop body.
next prev parent reply other threads:[~2014-12-19 5:34 UTC|newest]
Thread overview: 8+ messages / expand[flat|nested] mbox.gz Atom feed top
2014-12-19 3:45 Eric Cook
2014-12-19 5:16 ` Kurtis Rader
2014-12-19 5:34 ` Bart Schaefer [this message]
2014-12-19 5:48 ` Kurtis Rader
2014-12-19 5:51 ` Kurtis Rader
2014-12-19 6:05 ` Bart Schaefer
2014-12-19 11:15 ` Eric Cook
2014-12-19 9:35 ` Peter Stephenson
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