MIPS assembly code multiplier/divider. - Programmers Heaven

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# MIPS assembly code multiplier/divider.

Posts: 8Member
I have to write an unsigned 16-bit software implementation of a multiplier and a divider in MIPS assembly code. Two positive integers are read from the console. The program is supposed to multiply and divide the two integers, outputting the product, quotient and remainder.

[b]Here is an example run:[/b]
Enter an integer: 10
Enter an integer: 4
Product is 40
Quotient is 2
Remainder is 2

I can't use any MIPS multiply/divide instructions either.
CAN SOMEONE PLEASE HELP ME WITH THIS? I'M STILL A BEGINNER WITH MIPS ASSEMBLY LANGUAGE AND WOULD APPRECIATE ANY HELP. I know the process involves shifts, but I really need help with this project...please help.

## Comments

• Posts: 2,444Member
: I have to write an unsigned 16-bit software implementation of a
: multiplier and a divider in MIPS assembly code. Two positive
: integers are read from the console. The program is supposed to
: multiply and divide the two integers, outputting the product,
: quotient and remainder.
:
: [b]Here is an example run:[/b]
: Enter an integer: 10
: Enter an integer: 4
: Product is 40
: Quotient is 2
: Remainder is 2
:
: I can't use any MIPS multiply/divide instructions either.
: CAN SOMEONE PLEASE HELP ME WITH THIS? I'M STILL A BEGINNER WITH MIPS
: ASSEMBLY LANGUAGE AND WOULD APPRECIATE ANY HELP. I know the process
: involves shifts, but I really need help with this project...please
: help.
:

I don't know MIPS but the basics of multiplying and/or dividing two numbers using only addition and subtraction is:
Multiply: a * b -> Initialise a result variable to 0, create a loop from 1 to a, and in the loop add b to the result. Then at the end of the loop you'll have b+b+..+b for 'a' times, thus a*b
Division: a / b -> Initialise a counter variable to 0, create a loop that ends when a < b, and in this loop subtract b from a (a = a - b) and add one to counter. Then at the end of the loop, counter will hold the amount of times you've managed to subtract b from a }= a/b
(and incidentally, what's left in a is the remainder after division)
Best Regards,
Richard

The way I see it... Well, it's all pretty blurry
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