移位带来的除法问题

Question

我在使用MIPS实现除法算法时遇到了一些问题，我假设这与我移位和设置最低有效位的方式有关，但我不完全确定。

算法是这样的：

1)从余数寄存器中减去除数寄存器，并将结果放入余数寄存器。

2a)如果余数>=为0，则将商寄存器向左移位，将新的最右边的位设置为1

2b)如果余数<0，则将除数寄存器添加到余数寄存器，并将结果存储在余数寄存器中(以恢复余数之前的值)。将商寄存器向右移位，将最右边的位设置为0。

3)将除数寄存器右移一位

4)执行此n+1重复。因为我们做的是无符号的8位除法，所以我假设这意味着9次重复

编辑:为了简单起见，我只设置了$t0 = 12和$t1 = 5。

    .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.

Answer 1

对算法的描述在我看来有点不对劲。这是在C中工作的8位除法算法的样子：

unsigned char Q,R;
unsigned char N=12,D=5;
int i;

Q = 0;
R = 0;
for (i = 0; i < 8; i++) {
    R <<= 1;
    R |= (N & 0x80) >> 7;
    N <<= 1;
    Q <<= 1;
    if (R >= D) {
        R -= D;
        Q |= 1;
    }
}

将其转录成MIPS汇编应该不会太困难。