From: David Bergman (davidb_at_[hidden])
Date: 2003-10-21 12:52:25
I wrote:
> > (2) Even in your more syntactical view of things ("f(y) is
> always y"),
> > that argument is wrong
> >
> > z = any value, say 42 : f(z+1) = 42+1=43, (after previous
> > evaluation) z+1 = 43+1=44, not the same
Dave wrote:
> I didn't have such a syntactic view. And if you want to go
> down that road, whether or not z+1 == 44 really depends on
> how the language likes to print values ;->
try
std::cout << (f(z+1) == z+1) << std::endl;
PS
I know you understand these issues
DS
/David