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From: David Abrahams (dave_at_[hidden])
Date: 2003-10-21 12:23:22


"David Bergman" <davidb_at_[hidden]> writes:

> Actually, you are wrong (even though this has nothing to do with the purity
> of a language, since C++ is not pure...)
>
> The mistake you make is twofold:
>
> (1) The result of a computation is the canonical representation (read
> "numerical value" in this case)
>
> z = <n>: f(z) = <n>, which clearly varies as the value stored in 'z'
> varies... so, not the same at all

OK, OK, uncle! You win.

> (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

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 ;->

-- 
Dave Abrahams
Boost Consulting
www.boost-consulting.com

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