计算字符串输入中的单词数

Question

在以下代码中，32表示ASCII空格字符。我在移动字符串时面临一个问题：

.data
para: .asciiz " "
buffer: .space 250
ctr: .word 0
.text
.globl main
.ent main
main:addi $t3,$0,32
li $t8,1 
la $a0,para
li $v0,8
syscall
li $t0,0
loop:bge $t0,$a0,exit
add $t1,$a0,$t0
lb $t2,0($t1)
beq $t2,$t3,counter
addi $t0,$t0,1
j loop
counter:addi $t8,$t8,1
addi $t0,$t0,1
j loop 

exit:sw $t8,ctr
li $v0,10
syscall
.end main

Answer 1

我今天找到了答案，我在加载单个字符之前比较了数组：

.data
msg: .asciiz"Enter\n"
buffer: .space 250
ctr: .word 0

.text
.globl main
.ent main
main:addi $t3,$0,32    #$t3 holds 'space'
li $t8,1 

la $a0,msg
li $v0,4
syscall                        #$t8, is counter register

li $v0,8                          #input of string para
syscall

li $t0,0                           # i=0

loop:                    #till str[i]!=0
add $t1,$a0,$t0

lb $t2,0($t1)                   
beq $t2,$zero,exit
beq $t2,$t3,counter         #str[i]='  ' checks and jumps to 

addi $t0,$t0,1                   #increment i if false
j loop

counter: addi $t8,$t8,1     #increment counter when word 

addi $t0,$t0,1                  
j loop

exit: sw $t8,ctr
li $v0,10
syscall
.end main