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From: Bart Schaefer <schaefer@brasslantern.com>
Message-id: <141218213450.ZM28822@torch.brasslantern.com>
Date: Thu, 18 Dec 2014 21:34:50 -0800
In-reply-to: <54939F50.50102@gmx.com>
Comments: In reply to Eric Cook <llua@gmx.com>
 "Could someone clarify how math functions work?" (Dec 18, 10:45pm)
References: <54939F50.50102@gmx.com>
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Subject: Re: Could someone clarify how math functions work?
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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.