PIC Microcontoller Basic Math Division Methods

• 8 bit by 8bit to 24bit floating point AN575 format from Nikolai Golovchenko
• 16 bits by 8 from Lou Zher
• 16 bits by 8 with result as fraction rather than remainder by Nikolai Golovchenko
• div10byfxp7q8.htm Divide 10 bit integer using a fixed point divisor and result (7Q8 format)
• 24 bits by 8 from James Ashley Hillman
• 24 bits by 16 from Nikolai Golovchenko
• 24 bits by 16 from Andy David
• 24 bits by 24 from Tony Nixon
• 32 bits by 16 by Peter Hemsley +
• 32 bits by 16 using 16 bits by 16 by Nikolai Golovchenko
• 48 bits by 24 from Nikolai Golovchenko and Tony Kübek
• 48 bits by 24 from Andy Lee
• Binary Division by a Constant
• Tutorial
• Reciprocal
  • 32768/x by Nikolai Golovchenko
• Multiplication and/or division by a Constant
  • See: The PICList code generator by Nikolai Golovchenko
  • 16 bits by the constant 15 by Dmitry Kiryashov
  • 24 bits by the constant 6 by Hughe "can be used for any 8 bit constant divisor with 2 values changed."
  • 8 bits by the constant 3 by Andy Warren
• (8 bits * 256) / 8 bits by Nikolai Golovchenko
• Archive
  • PICList post "Challenge" detecting divisibility by 3
  • PICList post "16fxdu/8 divide routine"
• see also
  • http://home.netcom.com/~fastfwd/answers.html#PIC00094 from Andrew Warren: 16-bit by 8
  • http://home.netcom.com/~fastfwd/answers.html#PIC00064 from Andrew Warren: 8-bit by 4 PIC16C5x
  • http://www.myke.com/PICMicro/16bit.htm 16 bit by 8
  • http://www.myke.com/16bit.htm#div 16fxdu/8 divide routine
  • http://www.dontronics.com/see.html#division 16 bit by 8

Archive:

• List post "16 bit Mod 7 Math" +
• List post "Divide by 10"
• PICList post "coding challenge: divide by 7" +
• PICList post "Divide 24/8 routine" ASM embedded in C +
• PICList post "16fxdu/8 divide routine" +
• PICList post "16bit divide by 10" +
• PICList post "16/8 Bit Division with remainder"
• PICList post "Help needed with simple question"
• PICList post "24bit by 16bit Division" (actually 24 / 8)

See:

Questions:

Code:

• Comments:

• KILLspambillyman3-Remove- at KILLspamgmail.com

  Any reciprocal 1/n where n - 1 = 2^m (and m is an integer)can be represented as a geometric series starting with 1/(n-1) and having a ratio of -1/(n-1). This is convenient because division by n is just a matter of shifting. To expand on that point, dividing by another number, q, when q's factors are made up of the sequence described earlier (3, 5, 9, 17, 33...) and the sequence 2^m (2, 4, 8, 16, 32...) is a breeze.

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