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problem to understand how to divide two 16 bits number in 8051 microcontroller

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Fantin_G

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Hello everyone,

I have this project to code in 8051 : "Division of two 16 bit unsigned integers being in the internal memory, quotient and remainder should be stored".

I find a way to do it but there is a part of the program that i don't understand, I attach it.

I've noted which part i don't understand. By "don't understand", i mean that i'm not understanding why do we have to do all those calculations to find the quotient and the remainder.

I thank you in advance for your answer!
 

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  • 16 bit div.txt
    3 KB · Views: 392
I don't have time to work through that program line by line, but the basic principle of binary division in a computer is similar to one of the "long division" methods using pencil and paper.

You shift the divisor left by the number of digits you are working with - 1, and see if it can be subtracted from the dividend.
If it can, you add 1 (shifted by the same number of places) to the result.

Shift the divisor and "1" bit right one place and repeat, until the shift offset if zero, then you are done.

eg. just using four bits:

1100 / 0010
Result 00000000

00001100 - 0010000 (Initial shift 3 bits, "1" bit register set to 1000) Can't subtract, (and stay positive) so no change to the result.
Result 0000

Shift and repeat:
00001100 - 00001000 , 1 bit 0100; Subtraction works, add bit to result:
0100

Shift and repeat:
00000100 - 00000100 , 1 bit 0010; Subtraction works, add bit to result:
0110

Shift and repeat:
00000000 - 00000010 , 1 bit 0001; Subtraction does not work, result unchanged:
0110

The shift or bit offset in now zero, so operation complete. Anything still in the dividend register is the remainder.
(And 0110 is 1100 divided by two, 0010).

You example is using 16 bits, so two bytes or more for each value, resulting in more complex additions and subtraction tests etc., but seems to be pretty much the same concept.
 
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