still problems with ?:

[David Goodenough]

>From article <2632 at solo1.cs.vu.nl>, by maart at cs.vu.nl (Maarten Litmaath):
> chris at mimsy.UUCP (Chris Torek) writes:
> \...
> \But e1?e2:e3 *can* be turned into (e1&&e2 || !e1&&e3)
> 
> But what if e1 == *p++?

Simple:

((a = *p++) && e2 ) || (!a && e3)
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