我在使用 MIPS 实现除法算法时遇到了一些麻烦，我假设它与我移动和设置最低有效位的方式有关，但我并不完全确定。
算法是这样的：
1) 余数寄存器减去除数寄存器，将结果放入余数寄存器。
2a) 如果余数 >= 0，则将商寄存器左移，将新的最右边位设置为 1
2b) 如果余数<0，则将除数寄存器与余数寄存器相加，并将结果存入余数寄存器（恢复余数先前的值）。将商寄存器右移，将最右边的位设置为 0。
3) 将除数寄存器右移一位
4) 重复 n+1 次。由于我们正在进行无符号 8 位除法，我认为这意味着 9 次重复
编辑：为了简单起见，我只设置了 $t0 = 12 和 $t1 = 5。但是对于商，我得到 63，余数得到 0
.data
quotient_a: .asciiz "Quotient (a/b): "
remainder_a: .asciiz "Remainder (a/b):"
quotient_b: .asciiz "Quotient (b/a): "
remainder_b: .asciiz "Remainder (b/a):"
error: .asciiz "Sorry, divide-by-zero not allowed"
new_line: .asciiz "\n"
.text
main:
li $s0, 8 #8 bit
li $t0, 12
li $t1, 5
## Quotient A message
la $a0, quotient_a # load the addr of display_sum into $a0.
li $v0, 4 # 4 is the print_string syscall.
syscall # do the syscall.
##Computer the Quotient and remainder
## $t0 = a $t1 = b
## $t0 = dividend $t1 = divisor
li $t2, 0 ##Quotient
li $t3, 0 ##Remainder
li $t9, 0 ##i
li $s0, 9 ##count
loop:
sub $t3, $t3, $t1
blt $t3, 0, less_than
## >= 0
sll $t2, $t2, 1
addi $t2, $t2, 1
j cont
less_than:
add $t3, $t3, $t1
sll $t2, $t2, 1
cont:
srl $t1, $t1, 1
addi $t9, $t9, 1
bne $t9, $s0, loop
## print the quotient
move $a0, $t2
li $v0, 1 # load syscall print_int into $v0.
syscall # make the syscall.
## new line
la $a0, new_line # load the addr of new_line into $a0.
li $v0, 4 # 4 is the print_string syscall.
syscall
## Remainder A message
la $a0, remainder_a # load the addr of display_sum into $a0.
li $v0, 4 # 4 is the print_string syscall.
syscall # do the syscall.
## print the remainder
move $a0, $t3
li $v0, 1 # load syscall print_int into $v0.
syscall # make the syscall.
#Exit Program
li $v0, 10 # syscall code 10 is for exit.
syscall # make the syscall.