我的MIPS项目有问题。任务是将binary64（double）数乘以有符号整数而不使用浮点单位。它几乎运行良好，但对于某些数字发生错误（例如，float-123456789123456，int-1）。错误的意思是说，在第7位之后，我的结果似乎与正确的结果不同。我怀疑在程序中传输一些位可能会有问题（可能是在添加$ high的mantissa2和$ low of mantissa2的过程中）。我试图纠正它，但到目前为止我不知道该怎么做。请验证我的代码并在可能的情况下进行更正。
.data
text1: .asciiz "Enter double: "
text2: .asciiz "Enter integer: "
text3: .asciiz "Result: "
quest: .asciiz "\nIf you want to multiply enter 1, otherwise enter 0: "
num1a: .word 0 #multiplicand and result(sign exponent and first part of mantissa)
num1b: .word 0 #second part of the multiplicand and result(remaining part of mantissa)
num2: .word 0 #integer
.text
.globl input
input:
#print "Enter double: "
la $a0, text1
li $v0, 4
syscall
# saving input double into num1
li $v0, 7
syscall
swc1 $f0, num1b
swc1 $f1, num1a
#print "Enter integer: "
la $a0, text2
li $v0, 4
syscall
# saving input int into num2
li $v0, 5
syscall
sw $v0, num2
# loading data to registers
lw $t0, num1a
lw $t1, num1b
lw $t2, num2
#########################################################sign
sign:
move $t3, $t0
andi $t3, $t3, 0x80000000 #preserve sign, zero the rest
bgez $t2, extract #if less than zero we change the final sign and negate the value of integer
xori $t3, $t3, 0x80000000 #multiply signs (if integer is negative, then the sign is equal to $s0)
neg $t2, $t2 #absolute value of int
extract:
################################################checking for zero
or $t5, $t0, $t1 #if both part of double are equal to zero we skip all the calculation
beqz $t5, result_zero
beqz $t2, result_zero
###############################sign, exponent and mantissa
move $t7, $t0
andi $t7, $t7, 0x7FF00000 #extracting exponent to $t7
move $t8, $t0
andi $t8, $t8, 0x000FFFFF #extracting first part of mantissa
ori $t8, $t8, 0x00100000 #adding prefix one to mantissa
#remaining mantissa stays in register $t1
#########################################################
multiply:
########################multiplying mantissa part 1
multu $t8, $t2 #multiply mantissa1 by integer
mflo $t8 #low part of multiplication to $t8
mfhi $s1 #high part of multiplication to $s1
########################multiplying mantissa part 2
multu $t1, $t2 #mantissa part 2 multiplication
mflo $t1 #low part to $t1
mfhi $t0 #with overflow going to $t0
########################partial accumulation
addu $t8, $t8, $t0 #adding the high part of mantissa2 to result of low part of mantissa1
bgeu $t8, $t0, skip_add #if the result is less than any element we add 1 to mantissa1 high
addiu $s1, $s1, 1
######
skip_add:
bnez $s1, shift
bltu $t8, 0x00200000, result
shift: #else we shift 3 parts of mantissa and increment the the exponent
###extracting least significant bit of high mantissa1
sll $s2, $s1, 31 #copying least significant beat of $s1 to most significant bit in $s2
sll $t9, $t8, 31 #copying least significant beat of $s8 to most significant bit in $t9
######
srl $s1, $s1, 1 #shifting right mantisa part1 high
srl $t8, $t8, 1 #shifting right mantisa part1 low
or $t8, $t8, $s2 #copying least significant bit from mantissa1- high to most significant bit of mantissa1 low
srl $t1, $t1, 1 #shifting right mantisa part2
or $t1, $t1, $t9 #copying least significant bit from mantissa1 to most significant bit of mantissa2
######
addiu $t7, $t7, 0x00100000 #increment exponent by one
######
bnez $s1, shift #if mantissa1 high is greater than zero we continue
bgeu $t8, 0x00200000, shift #if mantissa1 low exceeds final mantissa space
result:
andi $t8, $t8, 0x000FFFFF #preserve mantissa, zero the rest(cut the prefix - one)
move $t0, $t3 #copy propoer sign
or $t0, $t0, $t7 #add exponent
or $t0, $t0, $t8 #add mantissa part1
b output
result_zero:
li $t0, 0
li $t1, 0
output:
sw $t0, num1a
sw $t1, num1b
#print "Result: "
la $a0, text3
li $v0, 4
syscall
lwc1 $f12, num1b
lwc1 $f13, num1a
#print double - the result
li $v0, 3
syscall
question:
la $a0, quest #Do you want to enter new numbers or finish?
li $v0, 4
syscall
li $v0, 5 #reads the answer (integer)
syscall
beq $v0, 1, input #if input =1, continue, if 0 finish, otherwise ask again
beqz $v0, fin
b question
fin:
li $v0, 10 #exit
syscall
我认为问题可能在本节（乘法）：
addu $t8, $t8, $t0 #adding the high part of mantissa2 to result of low part of mantissa1
bgeu $t8, $t0, skip_add #if the result is less than any element we add 1 to mantissa1 high
#addiu $s1, $s1, 1
######
skip_add:
bnez $s1, shift
bltu $t8, 0x00200000, result
添加过程可能导致执行。我尝试用注释（addiu）中的指令处理它。这意味着如果添加两个无符号数的结果小于其中一个，我们得到一个cary out并且必须添加1来注册$ s1，其中包含最重要的mantissa的一部分。它没有帮助。
答案 0 :(得分：1)
在我看来，这是正确的代码：
.data
text1: .asciiz "Enter double: "
text2: .asciiz "Enter integer: "
text3: .asciiz "Result: "
quest: .asciiz "\nIf you want to multiply enter 1, otherwise enter 0: "
num1a: .word 0 #multiplicand and result(sign exponent and first part of mantissa)
num1b: .word 0 #second part of the multiplicand and result(remaining part of mantissa)
num2: .word 0 #integer
.text
.globl input
input:
#print "Enter double: "
la $a0, text1
li $v0, 4
syscall
# saving input double into num1
li $v0, 7
syscall
swc1 $f0, num1b
swc1 $f1, num1a
#print "Enter integer: "
la $a0, text2
li $v0, 4
syscall
# saving input int into num2
li $v0, 5
syscall
sw $v0, num2
# loading data to registers
lw $t0, num1a
lw $t1, num1b
lw $t2, num2
#########################################################sign
sign:
move $t3, $t0
andi $t3, $t3, 0x80000000 #preserve sign, zero the rest
bgez $t2, extract #if less than zero we change the final sign and negate the value of integer
xori $t3, $t3, 0x80000000 #multiply signs (if integer is negative, then the sign is equal to $s0)
neg $t2, $t2 #absolute value of int
extract:
################################################checking for zero
or $t5, $t0, $t1 #if both part of double are equal to zero we skip all the calculation
beqz $t5, result_zero
beqz $t2, result_zero
###############################sign, exponent and mantissa
move $t7, $t0
andi $t7, $t7, 0x7FF00000 #extracting exponent to $t7
move $t8, $t0
andi $t8, $t8, 0x000FFFFF #extracting first part of mantissa
ori $t8, $t8, 0x00100000 #adding prefix one to mantissa
#remaining mantissa stays in register $t1
#########################################################
multiply:
########################multiplying mantissa part 1
multu $t8, $t2 #multiply mantissa1 by integer
mflo $t8 #low part of multiplication to $t8
mfhi $s1 #high part of multiplication to $s1
########################multiplying mantissa part 2
multu $t1, $t2 #mantissa part 2 multiplication
mflo $t1 #low part to $t1
mfhi $t0 #with overflow going to $t0
########################partial accumulation
addu $t8, $t8, $t0 #adding the high part of mantissa2 to result of low part of mantissa1
bgeu $t8, $t0, skip_add #if the result is less than any element we add 1 to mantissa1 high
addiu $s1, $s1, 1
######
skip_add:
bnez $s1, shift
bltu $t8, 0x00200000, result
shift: #else we shift 3 parts of mantissa and increment the the exponent
###extracting least significant bit of high mantissa1
sll $s2, $s1, 31 #copying least significant beat of $s1 to most significant bit in $s2
sll $t9, $t8, 31 #copying least significant beat of $s8 to most significant bit in $t9
######
srl $s1, $s1, 1 #shifting right mantisa part1 high
srl $t8, $t8, 1 #shifting right mantisa part1 low
or $t8, $t8, $s2 #copying least significant bit from mantissa1- high to most significant bit of mantissa1 low
srl $t1, $t1, 1 #shifting right mantisa part2
or $t1, $t1, $t9 #copying least significant bit from mantissa1 to most significant bit of mantissa2
######
addiu $t7, $t7, 0x00100000 #increment exponent by one
######
bnez $s1, shift #if mantissa1 high is greater than zero we continue
bgeu $t8, 0x00200000, shift #if mantissa1 low exceeds final mantissa space
result:
andi $t8, $t8, 0x000FFFFF #preserve mantissa, zero the rest(cut the prefix - one)
move $t0, $t3 #copy propoer sign
or $t0, $t0, $t7 #add exponent
or $t0, $t0, $t8 #add mantissa part1
b output
result_zero:
li $t0, 0
li $t1, 0
output:
sw $t0, num1a
sw $t1, num1b
#print "Result: "
la $a0, text3
li $v0, 4
syscall
lwc1 $f12, num1b
lwc1 $f13, num1a
#print double - the result
li $v0, 3
syscall
question:
la $a0, quest #Do you want to enter new numbers or finish?
li $v0, 4
syscall
li $v0, 5 #reads the answer (integer)
syscall
beq $v0, 1, input #if input =1, continue, if 0 finish, otherwise ask again
beqz $v0, fin
b question
fin:
li $v0, 10 #exit
syscall