有人可以帮我看看我的代码为什么会出现这样的错误? 我正在将十进制转换为二进制,这是我的代码
.data
msg1: .asciiz "Please insert value (A > 0) : "
msg2: .asciiz "Please insert the number system B you want to
convert to (2<=B<=10): "
#Above sting must be in one line
msg3: .asciiz "\nResult : "
.text
.globl main
main:
addi $s0,$zero,2
addi $s1,$zero,10
getA:
li $v0,4
la $a0,msg1
syscall
li $v0,5
syscall
blt $v0,$zero,getA
move $t0,$v0
getB:
li $v0,4
la $a0,msg2
syscall
li $v0,5
syscall
blt $v0,$s0,getB
bgt $v0,$s1,getB
add $t1,$zero,$v0
li $v0,4
la $a0,msg3
syscall
add $a0,$zero,$t0
add $a1,$zero,$t1
jal convert
li $v0,10
syscall
convert:
#a0=A
#a1=B
addi $sp,$sp,-16
sw $s3,12($sp) #counter,used to know
#how many times we will pop from stack
sw $s0,8($sp) #A
sw $s1,4($sp) #B
sw $ra,0($sp)
add $s0,$zero,$a0
add $s1,$zero,$a1
beqz $s0,end
div $t4,$s0,$s1 #t4=A/B
rem $t3,$s0,$s1 #t3=A%B
add $sp,$sp,-4
sw $t3,0($sp) #save t3
add $a0,$zero,$t4 #pass A/B
add $a1,$zero,$s1 #pass B
addi $s3,$s3,1
jal convert #call convert
end:
lw $ra,0($sp)
lw $s1,4($sp)
lw $s0,8($sp)
lw $s3,12($sp)
beqz $s3,done
lw $a0,16($sp)
li $v0,1
syscall
done:
addi $sp,$sp,20
jr $ra #return
当我运行此代码时,它在第13行表示语法错误,即li $v0,4
有人可以告诉我发生了什么吗?