Assembly MIPS汇编(MARS 4.5):浮点算术答案很奇怪
我正在制作一个MIPS程序来计算5.4xy-12.3y+18.23x-8.23y,其中x和y是来自控制台的输入。然而,结果太奇怪了。这是我的密码:Assembly MIPS汇编(MARS 4.5):浮点算术答案很奇怪,assembly,mips,Assembly,Mips,我正在制作一个MIPS程序来计算5.4xy-12.3y+18.23x-8.23y,其中x和y是来自控制台的输入。然而,结果太奇怪了。这是我的密码: .data promptX: .asciiz "Enter x: \n" promptY: .asciiz "Enter y: \n" result: .asciiz "Result: " first: .float 5.40 second: .float -12.30 third: .float 18
.data
promptX: .asciiz "Enter x: \n"
promptY: .asciiz "Enter y: \n"
result: .asciiz "Result: "
first: .float 5.40
second: .float -12.30
third: .float 18.23
fourth: .float -8.23
.text
# Print prompt to input x
li $v0, 4
la $a0, promptX
syscall
# Get x, store in f2
li $v0, 7
syscall
mov.d $f2, $f0
# Prompt input y
li $v0, 4
la $a0, promptY
syscall
# Get y, store in f4
li $v0, 7
syscall
mov.d $f4, $f0
# f6 = xy
mul.d $f6, $f2, $f4
# load 5.4 to f8
lwc1 $f8, first
# f6 = 5.4xy
mul.d $f6, $f6, $f8
# load -12.3 to f8
lwc1 $f8, second
# f8 = -12.3y
mul.d $f8, $f8, $f4
# f6 = 5.4xy - 12.3y
add.d $f6, $f6, $f8
# load 18.23 to f8
lwc1 $f8, third
# f8 = 18.23x
mul.d $f8, $f8, $f2
# f6 = 5.4xy - 12.3y + 18.23x
add.d $f6, $f6, $f8
# load -8.23 to f8
lwc1 $f8, fourth
# f6 = 5.4xy - 12.3y + 18.23x + (- 8.23)
add.d $f12, $f6, $f8
# Print answer
li $v0, 4
la $a0, result
syscall
li $v0, 2
syscall
# End program
li $v0, 10
syscall
当我输入x=2.13和y=2.13时,它返回1.26719E-10,其中应为28.90016。我反复检查了很多次代码,但仍然不知道为什么。非常感谢您的帮助。您正在混合单精度和双精度数字
.data部分中的系数是float
您将输入读取为双精度
使用lwc1作为浮点加载系数
将mul.d和add.d作为double执行计算
将结果打印为float
将所有内容切换到相同的精度,您的代码将正常工作。您正在混合单精度和双精度数字
.data部分中的系数是float
您将输入读取为双精度
使用lwc1作为浮点加载系数
将mul.d和add.d作为double执行计算
将结果打印为float
将所有内容切换到相同的精度,您的代码将正常工作。这将起作用:
.data
promptX: .asciiz "Enter x: \n"
promptY: .asciiz "Enter y: \n"
result: .asciiz "Result: "
first: .double 5.40
second: .double -12.30
third: .double 18.23
fourth: .double -8.23
.text
# Print prompt to input x
li $v0, 4
la $a0, promptX
syscall
# Get x, store in f2
li $v0, 7
syscall
mov.d $f2, $f0
# Prompt input y
li $v0, 4
la $a0, promptY
syscall
# Get y, store in f4
li $v0, 7
syscall
mov.d $f4, $f0
# f6 = xy
mul.d $f6, $f2, $f4
#mul.d $f2, $f4, $f6
# load 5.4 to f8
l.d $f8, first
# f6 = 5.4xy
#mul.d $f6, $f6, $f8
mul.d $f6, $f6, $f8
# load -12.3 to f8
l.d $f8, second
# f8 = -12.3y
mul.d $f8, $f8, $f4
# f6 = 5.4xy - 12.3y
add.d $f6, $f6, $f8
# load 18.23 to f8
l.d $f8, third
# f8 = 18.23x
mul.d $f8, $f8, $f2
# f6 = 5.4xy - 12.3y + 18.23x
add.d $f6, $f6, $f8
# load -8.23 to f8
l.d $f8, fourth
# f6 = 5.4xy - 12.3y + 18.23x + (- 8.23)
add.d $f12, $f6, $f8
# Print answer
li $v0, 4
la $a0, result
syscall
li $v0, 3
syscall
# End program
li $v0, 10
syscall
你把精确度搞混了。我把所有的东西都做成了双份。最后,您试图打印一个v0中有2个的浮点,您需要3个v0来打印一个双精度浮点
我注意到的另一个问题是将数据从内存加载到寄存器的方式。我把它们换成了l.d,数值加载正确 这将起作用:
.data
promptX: .asciiz "Enter x: \n"
promptY: .asciiz "Enter y: \n"
result: .asciiz "Result: "
first: .double 5.40
second: .double -12.30
third: .double 18.23
fourth: .double -8.23
.text
# Print prompt to input x
li $v0, 4
la $a0, promptX
syscall
# Get x, store in f2
li $v0, 7
syscall
mov.d $f2, $f0
# Prompt input y
li $v0, 4
la $a0, promptY
syscall
# Get y, store in f4
li $v0, 7
syscall
mov.d $f4, $f0
# f6 = xy
mul.d $f6, $f2, $f4
#mul.d $f2, $f4, $f6
# load 5.4 to f8
l.d $f8, first
# f6 = 5.4xy
#mul.d $f6, $f6, $f8
mul.d $f6, $f6, $f8
# load -12.3 to f8
l.d $f8, second
# f8 = -12.3y
mul.d $f8, $f8, $f4
# f6 = 5.4xy - 12.3y
add.d $f6, $f6, $f8
# load 18.23 to f8
l.d $f8, third
# f8 = 18.23x
mul.d $f8, $f8, $f2
# f6 = 5.4xy - 12.3y + 18.23x
add.d $f6, $f6, $f8
# load -8.23 to f8
l.d $f8, fourth
# f6 = 5.4xy - 12.3y + 18.23x + (- 8.23)
add.d $f12, $f6, $f8
# Print answer
li $v0, 4
la $a0, result
syscall
li $v0, 3
syscall
# End program
li $v0, 10
syscall
你把精确度搞混了。我把所有的东西都做成了双份。最后,您试图打印一个v0中有2个的浮点,您需要3个v0来打印一个双精度浮点
我注意到的另一个问题是将数据从内存加载到寄存器的方式。我把它们换成了l.d,数值加载正确