Compound Assignments

[Doug Gwyn]

In article <1991Apr11.183942.2195 at mccc.edu> pjh at mccc.edu (Pete Holsberg) writes:
>In article <15776 at smoke.brl.mil> gwyn at smoke.brl.mil (Doug Gwyn) writes:
>=In article <1991Apr8.174951.22448 at mccc.edu> pjh at mccc.edu (Pete Holsberg) writes:
>=>=	A compound assignment of the form E1 op= E2 differs from the
>=>=	simple assignment expression E1 = E1 op (E2) only in that the
>=>=	lvalue E1 is evaluated only once.
>=>The reference to E1 is ambiguous, as is the entire statement, IMHO. 
>=>*Which* E1 is evaluated "only once", the E1 op= E2 one or the other? 
>=>Does it follow that the remaining one is evaluated twice?  not at all?
>=You have GOT to be kidding -- there is nothing at all ambiguous about
>=the quoted specification.  It is an elegant way of expressing precisely
>=the semantics for op=.  I suggest you study it until enlightenment
>=suddenly dawns upon you.
>Nope!  [drivel deleted]

The point I was emphasizing is that you should be able to "grok" the
description with practically no effort.  If you're having difficulty,
you need to alter your thinking about it until you grok it.  Probably
your problem would be that you're trying to analyze the specification
at a more detailed level than necessary for understanding.

The description presupposes that you already understand what a simple
assignment expression means, and it builds on that knowledge to
explain what a compound assignment expression means.  It does that
by (a) rewriting into terms you should already understand and (b)
stating how the actual case differs from the rewrite.  I would think
that practically any native speaker of English who had already
assimilated the semantics of C through simple assignment expressions
should understand this specification for compound assignment with no
problem at all.  Certainly that's the way that I learned about the
semantics of compound assignment when I was first learning C, and it
was crystal clear to me then.