MIPS Decimal to Hex Conversion: How To Guide

[trouty323]

Hello. I was simply wondering if anyone has any idea on how to write a MIPS code to convert a decimal number to its hexadecimal equivalent. I have no idea on how I would go about this. Should I be trying to manipulate the ASCII values in some way? Or should I use some sort of shift/rotation method? I simply do not know the algorithm for completing this task. I'm completely lost.

 

[Grep]

You mean integers, I take it? Not floating point?

The wiki on hexadecimal goes through how to convert from base-10 to base-16, and should have what you want:

http://en.wikipedia.org/wiki/Hexadecimal

 

[trouty323]

The easiest way that I found was to have the user enter in an integer. I would then use a rotation instruction to rotate it 4 bits to the left. I would then mask those four bits with 0xf(1111). If those 4 digits were less than or equal to 9, I would add 48 to it to get the ASCII value. If it was 10 or above, I would add 55 to it. Here is my code in case anyone was wondering.

# Write a MIPS code that asks the user for decimal number
# Convert it to hex and print the result

	.data

prompt: .asciiz "Enter the decimal number to convert: "
ans: .asciiz "\nHexadecimal equivalent: "
result: .space 8

	.text
	.globl main

main:

	la $a0, prompt
	li $v0, 4
	syscall

	li $v0, 5
	syscall

	move $t2, $v0

	la $a0, ans
	li $v0, 4
	syscall

	li $t0, 8		        # counter
	la $t3, result		# where answer will be stored

Loop:

	beqz $t0, Exit		# branch to exit if counter is equal to zero
	rol $t2, $t2, 4		# rotate 4 bits to the left
	and $t4, $t2, 0xf	        # mask with 1111
	ble $t4, 9, Sum		# if less than or equal to nine, branch to sum
	addi $t4, $t4, 55	        # if greater than nine, add 55

	b End

	Sum:

		addi $t4, $t4, 48	# add 48 to result

End:

	sb $t4, 0($t3)		# store hex digit into result
	addi $t3, $t3, 1		# increment address counter
	addi $t0, $t0, -1		# decrement loop counter

j Loop

Exit:

	la $a0, result
	li $v0, 4
	syscall

	la $v0, 10
	syscall

 

[Mark44]

Your program seems to work correctly, although I have done only minimal testing. One thing you should change is your next to last line. Instead of la $v0, 10, you should have li $v0, 10.

 

[LudusRex]

I am familiar with the MIPS architecture and have some knowledge on coding in MIPS. Converting a decimal number to its hexadecimal equivalent can be done using a simple algorithm in MIPS. Here is a step-by-step guide on how to do it:

Step 1: Understand the Decimal and Hexadecimal Number Systems
Before we begin, it is important to understand the difference between decimal and hexadecimal number systems. Decimal numbers use a base of 10 and have 10 digits (0-9), while hexadecimal numbers use a base of 16 and have 16 digits (0-9 and A-F).

Step 2: Divide the Decimal Number by 16
To convert a decimal number to hexadecimal, we need to divide the decimal number by 16 repeatedly until the quotient becomes 0. The remainders obtained in each division will be the hexadecimal digits in reverse order.

Step 3: Convert Remainders to Hexadecimal Digits
Once we have the remainders, we need to convert them to their corresponding hexadecimal digits. For example, a remainder of 10 will be converted to A, 11 to B, and so on.

Step 4: Store the Hexadecimal Digits in Reverse Order
As mentioned earlier, the remainders will be the hexadecimal digits in reverse order. Therefore, we need to store them in a register in reverse order to get the correct hexadecimal equivalent of the decimal number.

Step 5: Print the Hexadecimal Equivalent
Finally, we can print the hexadecimal equivalent by converting the digits stored in the register to their corresponding ASCII values and printing them.

In summary, converting a decimal number to its hexadecimal equivalent in MIPS involves dividing the number by 16, converting the remainders to hexadecimal digits, storing them in reverse order, and then printing the result. I hope this guide helps you in writing your MIPS code for decimal to hexadecimal conversion.