From: Will Bryant (will_at_[hidden])
Date: 2005-09-22 04:14:46
Gennadiy Rozental wrote:
>BOOST_CHECK_CLOSE algorithm involves division (it compared relative distance
>with tolerance). |a-b|/|a| will always produce 0 for integral types. Which
>is less then any tolerance you could specify.
>
>
But we're comparing it to a 0-100 percentage number, so why can't we
just multiply before the division?
-- Will Bryant http://carcino.gen.nz/ will_at_[hidden] +64 21 655 443