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  3. In C#: double d = 0.4 - 1.2; What is d?

In C#: double d = 0.4 - 1.2; What is d?

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  • I Ivor S Sargoytchev

    Well? What is d equal to? Ivor S. Sargoytchev Dundas Software

    D Offline
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    Daniel Turini
    wrote on last edited by
    #2

    It will return the same as any IEEE floating point software will: rounding errors. http://www.statsdirect.com/discussion/_disc1/0000000d.htm[^]

    // Quantum sort algorithm implementation
    while (!sorted)
    ;

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    • I Ivor S Sargoytchev

      Well? What is d equal to? Ivor S. Sargoytchev Dundas Software

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      leppie
      wrote on last edited by
      #3

      Thru a microscope, everything looks big. Believe me that is -0.8. Try Console.Write(). leppie::AllocCPArticle(Generic DFA State Machine for .NET);

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      • I Ivor S Sargoytchev

        Well? What is d equal to? Ivor S. Sargoytchev Dundas Software

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        Tim Smith
        wrote on last edited by
        #4

        Integer Math != IEEE Math Tim Smith I'm going to patent thought. I have yet to see any prior art.

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        • L leppie

          Thru a microscope, everything looks big. Believe me that is -0.8. Try Console.Write(). leppie::AllocCPArticle(Generic DFA State Machine for .NET);

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          Ivor S Sargoytchev
          wrote on last edited by
          #5

          It is not since "if (d == -0.8)" will fail. Ivor S. Sargoytchev Dundas Software

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          • D Daniel Turini

            It will return the same as any IEEE floating point software will: rounding errors. http://www.statsdirect.com/discussion/_disc1/0000000d.htm[^]

            // Quantum sort algorithm implementation
            while (!sorted)
            ;

            N Offline
            N Offline
            Nemanja Trifunovic
            wrote on last edited by
            #6

            Hehehe, I ran into this problem 15 years ago while writing a Fortran program.

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            • N Nemanja Trifunovic

              Hehehe, I ran into this problem 15 years ago while writing a Fortran program.

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              Ivor S Sargoytchev
              wrote on last edited by
              #7

              I guess I should use decimals. Ivor S. Sargoytchev Dundas Software

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              • I Ivor S Sargoytchev

                I guess I should use decimals. Ivor S. Sargoytchev Dundas Software

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                Daniel Turini
                wrote on last edited by
                #8

                Ivor S. Sargoytchev wrote: I guess I should use decimals. Decimals are very, very slow. Depending on what you're doing, there are techniques to workaround most IEEE rounding problems.

                // Quantum sort algorithm implementation
                while (!sorted)
                ;

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                • I Ivor S Sargoytchev

                  It is not since "if (d == -0.8)" will fail. Ivor S. Sargoytchev Dundas Software

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                  Daniel Turini
                  wrote on last edited by
                  #9

                  Yes, but you should compare it like:

                  if (Math.Abs(d - 0.8) < Double.Epsilon)

                  And everything looks fine. See Double.Epsilon docs on MSDN.

                  // Quantum sort algorithm implementation
                  while (!sorted)
                  ;

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                  • D Daniel Turini

                    Yes, but you should compare it like:

                    if (Math.Abs(d - 0.8) < Double.Epsilon)

                    And everything looks fine. See Double.Epsilon docs on MSDN.

                    // Quantum sort algorithm implementation
                    while (!sorted)
                    ;

                    I Offline
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                    Ivor S Sargoytchev
                    wrote on last edited by
                    #10

                    Daniel, This is not true. Double.Epsilon is a very very very small number: Double.Epsilon = 4.94065645841247E-324 But in C# I get 0.4 - 1.2 = -0.79999999999999993. As you can see the deviation from -0.8 is hundreds of times bugger than 4.94065645841247E-324. Ivor S. Sargoytchev Dundas Software

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                    • I Ivor S Sargoytchev

                      It is not since "if (d == -0.8)" will fail. Ivor S. Sargoytchev Dundas Software

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                      Chris Maunder
                      wrote on last edited by
                      #11

                      Ivor S. Sargoytchev wrote: if (d == -0.8) You're comparing floating point values for equality?? :omg: cheers, Chris Maunder

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                      • C Chris Maunder

                        Ivor S. Sargoytchev wrote: if (d == -0.8) You're comparing floating point values for equality?? :omg: cheers, Chris Maunder

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                        Ivor S Sargoytchev
                        wrote on last edited by
                        #12

                        Chris, With respect, some constructive criticism please. Ivor S. Sargoytchev Dundas Software

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                        • I Ivor S Sargoytchev

                          Chris, With respect, some constructive criticism please. Ivor S. Sargoytchev Dundas Software

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                          peterchen
                          wrote on last edited by
                          #13

                          ok, you asked for it... It is *well known* (;P) that floating point numbers have a limited resolution (around 7 digits for single, 15 for double*), and comparing the result of an arithmetic operation using "==" is prone to err. if you know the order of magnitude of one comparand, or at least you know that it's "far enough" from zero, use the abs(a-b) < 1e-6*(abs of known comparand) comparison. Otherwise it gets a little bit more tricky. *) don't quote me on the numbers, look up the IEEE definitions for the format


                          "Der Geist des Kriegers ist erwacht / Ich hab die Macht" StS
                          sighist | Agile Programming | doxygen

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                          • I Ivor S Sargoytchev

                            Chris, With respect, some constructive criticism please. Ivor S. Sargoytchev Dundas Software

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                            Chris Maunder
                            wrote on last edited by
                            #14

                            The value represented in a floating point variable is only an approximation to a value, not an exact representation. Because of this you cannot assume that something like 2.0 == 4.0 * 0.5 will return true. When comparing floating point values you need to estimate how close is close enough then do a fuzzy comparision (something like ABS(A-B) < Epsilon) cheers, Chris Maunder

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                            • C Chris Maunder

                              The value represented in a floating point variable is only an approximation to a value, not an exact representation. Because of this you cannot assume that something like 2.0 == 4.0 * 0.5 will return true. When comparing floating point values you need to estimate how close is close enough then do a fuzzy comparision (something like ABS(A-B) < Epsilon) cheers, Chris Maunder

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                              Lost User
                              wrote on last edited by
                              #15

                              Chris Maunder wrote: The value represented in a floating point variable is only an approximation to a value, not an exact representation. Because of this you cannot assume that something like 2.0 == 4.0 * 0.5 will return true. So this is how you count your beer consumption isn't? Especially when Kiwi and Aussie mates are involved. ;P Michael Martin Australia "I personally love it because I can get as down and dirty as I want on the backend, while also being able to dabble with fun scripting and presentation games on the front end." - Chris Maunder 15/07/2002

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                              • I Ivor S Sargoytchev

                                Well? What is d equal to? Ivor S. Sargoytchev Dundas Software

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                                Brit
                                wrote on last edited by
                                #16

                                If it's anything like C++, then d = -0.79999999999999993 ------------------------------------------ The ousted but stubbornly non-dead leader reportedly released an audiotape this weekend, ending by calling on Iraqis to, quote, "resist the occupation in any way you can, from writing on walls, to boycotting, to demonstrating and taking up arms." adding, "you know, pretty much anything I used to kill you for."

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                                • C Chris Maunder

                                  The value represented in a floating point variable is only an approximation to a value, not an exact representation. Because of this you cannot assume that something like 2.0 == 4.0 * 0.5 will return true. When comparing floating point values you need to estimate how close is close enough then do a fuzzy comparision (something like ABS(A-B) < Epsilon) cheers, Chris Maunder

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                                  Michael Dunn
                                  wrote on last edited by
                                  #17

                                  Chris Maunder wrote: 2.0 == 4.0 * 0.5 Won't that actually be true because all those numbers are powers of 2? :cool: --Mike-- Ericahist | Homepage | RightClick-Encrypt | 1ClickPicGrabber Latest art~!@#2rDFA#@(#*%$Rfa39f3fqwf--= NO CARRIER

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                                  • T Tim Smith

                                    Integer Math != IEEE Math Tim Smith I'm going to patent thought. I have yet to see any prior art.

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                                    Michael Dunn
                                    wrote on last edited by
                                    #18

                                    That reminds me of the old C= BASIC, there was some number (I'll just use 42 as an example) where you could type:

                                    PRINT 42

                                    and the response would be:

                                    42.0001

                                    :wtf: --Mike-- Ericahist | Homepage | RightClick-Encrypt | 1ClickPicGrabber Latest art~!@#2rDFA#@(#*%$Rfa39f3fqwf--= NO CARRIER

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                                    • M Michael Dunn

                                      Chris Maunder wrote: 2.0 == 4.0 * 0.5 Won't that actually be true because all those numbers are powers of 2? :cool: --Mike-- Ericahist | Homepage | RightClick-Encrypt | 1ClickPicGrabber Latest art~!@#2rDFA#@(#*%$Rfa39f3fqwf--= NO CARRIER

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                                      John M Drescher
                                      wrote on last edited by
                                      #19

                                      Yes it should... John

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                                      • I Ivor S Sargoytchev

                                        Well? What is d equal to? Ivor S. Sargoytchev Dundas Software

                                        N Offline
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                                        Navin
                                        wrote on last edited by
                                        #20

                                        Ivor S. Sargoytchev wrote: Dundas Software THe only way you're going to get an answer in this crowd, when posting in the Lounge, is if you offer to hook us up with those women from the Summer Sale ads. :-D "When a man sits with a pretty girl for an hour, it seems like a minute. But let him sit on a hot stove for a minute and it's longer than any hour. That's relativity." - Albert Einstein

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                                        • I Ivor S Sargoytchev

                                          Daniel, This is not true. Double.Epsilon is a very very very small number: Double.Epsilon = 4.94065645841247E-324 But in C# I get 0.4 - 1.2 = -0.79999999999999993. As you can see the deviation from -0.8 is hundreds of times bugger than 4.94065645841247E-324. Ivor S. Sargoytchev Dundas Software

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                                          Rodolfo Lima
                                          wrote on last edited by
                                          #21

                                          I think that what he meant was something like: if(abs(0.4-1.2)-0.8) < Double.Epsilon) which is the proper way to compare floating point numbers, i guess.

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