計算字符串輸入的字數

Question

在下面的代碼中，32是ASCII空格字符。我現在面臨一個問題，在字符串中移動：

.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

我找到了答案，我今天被加載單個字符前陣比較：

.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