Question

当我等待减法的结果出现时出现问题，但它显示我0.不是x-a，正如我希望的那样。

.data


strin: .asciiz "type two integers\n\n\n" 

strout: .asciiz "the result of the substract is:"

a: .word 0

x: .word 1



.text

main:

li $v0, 4                                                                          
la $a0,strin                                                                 
syscall  

li $v0, 5
syscall
sw $t0,a

li $v1,5
syscall
sw $t1,x


sub $t1,$t1,$t0                                      


li $v0, 4                                                                          
la $a0,strout                                                                                   
syscall

move $a0, $t1                   
li $v0, 1                        
syscall

li $v0,10                                               
syscall

Answer 1

更好的方法是将整数作为输入读取，然后进行计算。将值存储在内存中然后将其读取为不是一种实用的方法如果您不想将数字作为输入读取，我建议您使用li指令将特定数值加载到寄存器中，然后继续进行计算。

.data
strin: .asciiz "type two integers\n"
strout: .asciiz "the result of the substract is:"

.text
.globl main
main:

li $v0, 4
la $a0,strin                                                                 
syscall

#changed
li $v0, 5        #Read the first integer
syscall
move $t0,$v0     #move the value you read to $t0 register

#changed
li $v0, 5       #Read the second integer
syscall
move $t1,$v0    #move the value you read to $t1 register

sub $s0,$t0,$t1    #make substraction between $t0 and $t1 and store the        
                   #result in $s0 register       
li $v0, 4
la $a0,strout                                                                                   
syscall

li $v0, 1
move $a0, $s0
syscall

li $v0,10
syscall

在Mips中减去

1 个答案: