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Data conversion,a simple but strange problem,please help!Hi,
We all know that the return value of Math.Log(8,2) is 3,but how about (int)Math.Log(8,2)? On my machine,the return value of (int)Math.Log(8,2) is strange enough! it's not 3 but 2 ! I've tested other values such as (int)Math.Log(16,2),(int)Math.Log(4,2),(int)Math.Log(32,2)...et, they all return the same value as their counterpart Math.Log() methods,only (int)Math.Log(8,2) is incorrect,is this a .net bug ? I've also tested Math.Floor(Math.Log(8,2)),its return value is same as (int)Math.Log(8,2),but Math.Ceiling(Math.Log(8,2)) and int.Parse(Math.Log(8,2).ToString()) returns the correct value of 3! Dose anyone know what caused this strange behavior? thanks in advance! "Webdiyer" <webdi***@hotmail.com> wrote in message No it's not. Math.Log(8,2) is a double precision floating point value news:OJt2zJfOGHA.668@TK2MSFTNGP11.phx.gbl... > Hi, > > We all know that the return value of Math.Log(8,2) is 3, guaranteed only to be pretty darn close to 3. It's like the add for the new Delta shaver: It's not just close, it's Eplsilon-close! In particular (3 - double.Epsilon) < Math.Log(8,2) < (3 + double.Epsilon). When you convert it to an integral type, you should explicitly apply whatever rounding behavior you want. Or else perform your arithmetic on integers only. David Thank you for your quick reply,I've solved this problem by first converting
Math.Log(8,2) to decimal value and then converting this value to int,this way the result is correct. Show quote "David Browne" <davidbaxterbrowne no potted m***@hotmail.com> wrote in message news:OHtxQ%23gOGHA.3888@TK2MSFTNGP12.phx.gbl... > > "Webdiyer" <webdi***@hotmail.com> wrote in message > news:OJt2zJfOGHA.668@TK2MSFTNGP11.phx.gbl... >> Hi, >> >> We all know that the return value of Math.Log(8,2) is 3, > > No it's not. Math.Log(8,2) is a double precision floating point value > guaranteed only to be pretty darn close to 3. > > It's like the add for the new Delta shaver: > It's not just close, it's Eplsilon-close! > > In particular (3 - double.Epsilon) < Math.Log(8,2) < (3 + > double.Epsilon). > > When you convert it to an integral type, you should explicitly apply > whatever rounding behavior you want. Or else perform your arithmetic on > integers only. > > David >  |
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