[EMAIL PROTECTED] (Dag-Erling Sm�rgrav) writes:
> David Schultz <[EMAIL PROTECTED]> writes:
> > When the line is there, the compiler is probably smart enough to
> > realize that 'x=y; y=x' is (usually) a no-op, so it optimizes away
> > both statements.
> Wrong.  The compiler is free to optimize away the second statement
> provided that neither x nor y is declared volatile, but it cannot
> optimize away the first statement.

I should add: unless it can determine with absolute certainty that x
is not referenced later.

DES
-- 
Dag-Erling Sm�rgrav - [EMAIL PROTECTED]
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