On 11/02/2009 12:37 PM, "Luke the Hiesterman" <[email protected]> wrote:

> Would it really be that much faster? I don't know exactly how pow() is
> implemented, but I assume it's basically just a loop of
> multiplications, in which case it would basically be the same as x*x
> in this case, since it would exit after the first iteration....
> 
> Luke

pow takes a double for the parameter, not an integer, so pow doesn't
necessarily perform a loop (it may effectively be a single machine
instruction).  Also, in early Rosetta releases of Mac OS X, we found that
doing pow(10,3) would produce 1000.00...01 rather than 1000, which was
definitely a bug, but it illustrates the difference.


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