ftp.cs.arizona.edu/ftp.cs.arizona.edu.tar/ftp.cs.arizona.edu/icon/newsgrp/group93b.txt/000105_icon-group-sender _Wed May 19 18:19:10 1993.msg

Internet Message Format | 1993-06-16 | 2KB

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Date: 19 May 93 18:19:10 GMT
From: agate!howland.reston.ans.net!ux1.cso.uiuc.edu!uchinews!ellis![e-mail]  (Richard L. Goerwitz)
Organization: University of Chicago
Subject: Re: A bug? Or a feature?
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John Nall writes that the Icon manual says:
>
>    X1 op:= X2
>      "performs the operation X1 op X2 and assigns the result to X1
>
>   ...To wit: it actually:
>
>    "performs the operation X1 op (X2) and assigns the result to X1"

I see what you mean.  The problem is with how you define X1 and X2.

Here's the scoup:  Think of X1 and X2 as expression 1 and expression 2.
How do we group expressions?  By operator precedence and newlines.  What
is the precedence of <:=, for example?  It is the same as any assignment
operator's.  What is the precedence of >, say?  It is the same as that
of any comparison operator.

Now we can answer the question of how the following expression groups:

>        i <:= j > k

Since comparison has a higher precedence than assignment, we group (j > k)
as one expression.  Hence the correct grouping of the entire assignment
expression is:

    i <:= (j > k)

Ergo i is expression 1, and (j > k) is expression 2.  The augmented assign-
ment operator adds the result of expression 2 to expression 1, and assigns
the result to expression 1 (if, in fact, expression 1 produces a variable).

So the Icon book is, in fact, correct, using its own definitions.  op:= is
not a macro; it is a regular assignment operator.  You just have to figure
out how the surrounding expressions group.

Note:

What your comments indicate is that the Icon book's explanation, while cor-
rect, may not be intuitive to everyone.  Good question.

--

   -Richard L. Goerwitz              [e-mail]
   [e-mail]          rutgers!oddjob!ellis!goer