MIPS Decimal to Hex Conversion: How To Guide

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SUMMARY

This discussion provides a comprehensive guide on writing MIPS assembly code to convert a decimal number to its hexadecimal equivalent. The key algorithm involves rotating the decimal number 4 bits to the left, masking the result with 0xf, and adjusting the value to obtain the corresponding ASCII character. The provided MIPS code effectively demonstrates this process, including user input and output of the hexadecimal result. A correction is noted regarding the use of 'li $v0, 10' instead of 'la $v0, 10' for program termination.

PREREQUISITES
  • Understanding of MIPS assembly language syntax and structure
  • Familiarity with bitwise operations, specifically rotation and masking
  • Knowledge of ASCII character encoding for numerical values
  • Basic experience with user input and output in MIPS
NEXT STEPS
  • Study MIPS assembly language documentation for further insights on system calls
  • Learn about bitwise operations in depth, focusing on rotation and masking techniques
  • Explore ASCII encoding and its application in data representation
  • Investigate error handling in MIPS programs to improve robustness
USEFUL FOR

Students and developers working with MIPS assembly language, particularly those interested in low-level programming and numerical conversions. This discussion is also beneficial for anyone seeking to enhance their understanding of bitwise operations and ASCII manipulation.

trouty323
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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.
 
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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
 
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.

Code:
# 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
 
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.
 
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