Re: [Boost-users] BOOST_CHECK_CLOSE with integral types

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@core-dev.co.nz +64 21 655 443