mips司实施

Question

我正在尝试在MIPS中实现除法算法，我应该执行以下操作：

1. 余数和商位于同一个寄存器中，上半部分是余数，下半部分是商，该寄存器被初始化为被除数的值。
2. 将余数寄存器左移 1 位
3. ，然后从余数寄存器的余数部分减去除数，仅
4. 检查 rem 是否小于 0，如果 rem < 0 0 则 rem = rem + 除数，如果 rem > 0，然后只放入第一位（最低有效位为1）
5. 继续执行n次，其中n是除数的宽度，在我的例子中它是6位宽，所以循环是6次

这里是代码到目前为止，我已经写过，我试图首先对正操作数进行操作，然后想对签名操作数进行操作

# A program that divides two integers according to the approach described in Fig. 3.12

.data   # Data declaration section

.text

main:       # Start of code section 
li $s2, 4 # dividend, will als0 be used as the remainder register which will hold remainder and quotient initialized to dividend
li $s3, 2 # divisor

li $s4, 2 # another register to hold the dividend shifted 6 bits, for adding and subtracting dividend from remainder
sll $s4, $s4, 6

li $t0, 0 # counter of the loop

LOOP:   sll $s2, $s2, 1 # shift the remainder regitser by 1 bit to right
sub $s2, $s2, $s4 # subtract divisor from remainder part of the remainder register

slt $t1, $s2, $zero # to check if rem < 0
beq $t1, $zero, MORE # if rem no < 0 then branch to MORE label
nop

add $s2, $s2, $s4 # if rem < 0, to add the divisor to the remainder part of the remainder register

j LOOP # jump back to the loop
nop

MORE:   # if rem > 0, then do arithmetic right shift and place 1 as the 0th position
rol $s2, $s2, 1 # rotate the number to the left by 1 bit which is arithmetic right shift

j LOOP # jump back to loop
nop

addi $t0, $t0, 1 # adding 1 to the counter of the loop
slti $t1, $t0, 6 # checking if the loop condition is working or not
bne $t1, $zero, LOOP
nop

add $a0, $zero, $s2 # putting the result in regitser a0

li $v0, 1 # printing out the result
syscall

# END OF PROGRAM

有人可以检查我的代码，并告诉我哪里出错了。 谢谢

Answer 1

请参阅此问题并回答 C 中的参考代码。有符号整数的直接除法相当复杂。如果您足够勇敢，请查阅布斯除法算法。