Assembly Case语句代码在Mars模拟器中运行，但在QtSpim中出错
正如标题所提到的，我已经在Mars MIPS模拟器中编写了这段代码，在那里组装和运行代码工作正常，没有错误。当我将同一个文件带到QtSpim并在那里运行时，会出现以下错误： 似乎QtSpim在我调用Assembly Case语句代码在Mars模拟器中运行，但在QtSpim中出错,assembly,mips,Assembly,Mips,正如标题所提到的，我已经在Mars MIPS模拟器中编写了这段代码，在那里组装和运行代码工作正常，没有错误。当我将同一个文件带到QtSpim并在那里运行时，会出现以下错误： 似乎QtSpim在我调用CASE的标签地址时遇到了问题，我没有足够的经验来了解原因。这是我的密码： .data jumptable: .word isD, is1, is2, is3, is4 currentvalue: .asciiz "Returned : i = " #Used for result outpu
CASE.data
jumptable: .word isD, is1, is2, is3, is4
currentvalue: .asciiz "Returned : i = " #Used for result output
space: .asciiz " \n" #Used for result output
.text
MAIN:
li $a0, 0 #testing use of the argument register toholdinput
jal CASE #set $ra and go to the case procedure
li $v0, 4 #load syscall 4 to print text
la $a0, currentvalue #load the string value of currentvalue into $a0
syscall #print out currentvalue
li $v0, 1 #load syscall 1 to print integer
move $a0, $t3 #move the contents of $t3 to $a0
syscall #print the integer in $a0
li $v0, 10 #load syscall 10 to exit program
syscall #exit program
CASE:
la $t3, ($a0) #load 'i' into a temporary value since we will be using $a0 for syscalls shortly
blt $a0, 1, isD #jump to default if i less than 1
bgt $a0, 4, isD #jump to default if i greater than 4
sll $t0, $a0, 2 #multiplies 'i' by 2^2
la $t1, jumptable #load the base address of the jump table into $t1
add $t1, $t1, $t0 #$t1 = $t1(base address of table) + $t0('i' * 4)
lw $t2, ($t1) #$t2 is loaded with the address of label found in $t1
jr $t2
is1:
addi $t3, $t3, 1 #increment 'i' by 1
jr $ra #jump back to the return address in main
is2:
addi $t3, $t3, 2 #increment 'i' by 2
jr $ra #jump back to the return address in main
is3:
addi $t3, $t3, 3 #increment 'i' by 3
jr $ra #jump back to the return address in main
is4:
addi $t3, $t3, 4 #increment 'i' by 4
jr $ra #jump back to the return address in main
isD:
li $t0, 0 #set 'i' to 0
jr $ra #jump back to the return address in main
好吧，我想我已经明白了。（警告：我只有火星） 在错误消息中，错误来自第188行，因此这不是程序的一部分。它是
spimmainmainmarsspimMAIN:mars.text更新：
另一方面，如果你有机会的话，我现在正在学习MIPS，根据我的评论，我的代码中的每一件事情都是正确的吗 我做了我版本的代码审查，这意味着我对源代码进行了注释，并对其进行了清理，直到我理解了所发生的事情。对你来说，这只是编辑一些评论。我确实在这个过程中发现了一个小的[一行]错误，我在下面的注释代码中修复了它 你在舞台上的评论做得很好。一些OPs在没有任何评论的情况下发布asm。好的评论[用任何语言]都应该表现出意图。也就是说，您希望您的程序在给定的点上做什么（“什么/为什么”）。实际的asm说明是“如何” 正如您所做的，我建议您对大多数asm说明使用侧边栏注释。我还建议在函数上方添加注释块 我清理了你的评论，就像我自己清理一样，因为其中一些评论如下：
x = 23; // set the value of x to twenty three
i<1i+=3case .data
jumptable: .word isD,is1,is2,is3,is4
prompt: .asciiz "Enter case input argument: "
currentvalue: .asciiz "Returned : i = " # Used for result output
space: .asciiz " \n" # Used for result output
.text
.globl main
main:
# prompt user for test value
li $v0,4 # syscall: print string
la $a0,prompt
syscall
# get user test value from stdin
li $v0,5 # syscall: read integer
syscall
move $a0,$v0 # move to argument for case
# LEARNER: testing use of the arg reg to hold input
###li $a0,0 # test value for case function
# LEARNER: set $ra and go to the case procedure (i.e. we are calling the
# case function)
jal case # calculate [whatever case calculates ;-)]
# print the prefix string
li $v0,4 # syscall: print string
# LEARNER: address of currentvalue into $a0
la $a0,currentvalue
syscall
# print the value the 'case' function returned
li $v0,1 # syscall: print integer
# LEARNER: move the contents of $t3 to $a0
move $a0,$t3 # get value to reg for syscall
syscall
li $v0,10 # syscall: exit program
syscall
# case -- does what?
#
# RETURNS: what does it return?
# t3 -- return value
# FIXME: to follow ABI, this should be v0 but okay for small internal
# functions like this one.
#
# arguments:
# a0 -- contains 'i' (FIXME: what is 'i'?)
#
# internal register usage (i.e. what registers get clobbered):
# t0 -- offset into jump table
# t1 -- address of jump table
# t2 -- internal jump address
case:
move $t3,$a0 # save 'i' to our work/return register
blt $a0,1,isD # jump to default if i < 1
bgt $a0,4,isD # jump to default if i > 4
# LEARNER: multiply 'i' by 2^2 (4) --> i <<= 2
sll $t0,$a0,2 # get the offset into the jump table
la $t1,jumptable # base address of the jump table
# LEARNER: $t1 = $t1(base address of table) + $t0('i' * 4)
add $t1,$t1,$t0 # add offset to base address
# LEARNER: $t2 is loaded with the address of label found in $t1
lw $t2,($t1) # jump_address = jumptable[i]
jr $t2 # jump to desired label
is1:
addi $t3,$t3,1 # i += 1
jr $ra # return to caller
is2:
addi $t3,$t3,2 # i += 2
jr $ra # return to caller
is3:
addi $t3,$t3,3 # i += 3
jr $ra # return to caller
is4:
addi $t3,$t3,4 # i += 4
jr $ra # return to caller
isD:
# BUG: this should be t3
###li $t0,0 # i = 0
li $t3,0 # i = 0
jr $ra # return to caller
.data
跳转表：。字isD、is1、is2、is3、is4
提示：.asciiz“输入案例输入参数：”
currentvalue:.asciiz“返回：i=“#用于结果输出
空格：.asciiz“\n”#用于结果输出
.文本
格洛博梅因酒店
主要内容：
#提示用户输入测试值
li$v0,4#系统调用：打印字符串
la$a0，提示
系统调用
#从stdin获取用户测试值
li$v0,5#系统调用：读取整数
系统调用
移动$a0，$v0#移动到案例的参数
#学员：测试使用arg reg来保持输入
###li$a0,0#机箱功能的测试值
#学员：设置$ra并进入案例程序（即，我们调用
#案例功能）
日航案例#计算[任何案例计算；-）]
#打印前缀字符串
li$v0,4#系统调用：打印字符串
#学习者：将currentvalue转换为$a0的地址
la$a0，当前值
系统调用
#打印“case”函数返回的值
li$v0,1#系统调用：打印整数
#学员：将$t3的内容移动到$a0
将$a0、$t3#获取值以注册系统调用
系统调用
li$v0,10#系统调用：退出程序
系统调用
#凯斯，做什么？
#
#返回：返回什么？
#t3——返回值
#修正：为了遵循ABI，这应该是v0，但对于小的内部
#像这样的功能。
#
#论据：
#a0——包含“i”（修正：什么是“i”
#
#内部寄存器使用（即，哪些寄存器被阻塞）：
#t0——跳转表中的偏移量
#t1——跳转表的地址
#t2——内部跳转地址
案例：
将$t3、$a0#保存“i”到我们的工作/回报登记簿
blt$a0,1，isD#如果i<1，则跳转为默认值
bgt$a0,4，isD#如果i>4，则跳转到默认值
#学习者：将“i”乘以2^2（4）-->我怀疑SPIM区分大小写，并尝试跳到main
，而您的代码只定义main
。哇，请注意，我将main和case分别改为main和case，瞧，我得到了输出。非常感谢。这是有道理的，我刚刚重命名了标签，它在Spim中对我有效，但到重命名时还没有在Mars中尝试过。另一方面，如果你有机会，我现在正在学习MIPS，根据我的评论，我的代码中的每件事看起来都很好吗？非常感谢你。我真的很感激！
x = 23; // set the value of x to twenty three
.data
jumptable: .word isD,is1,is2,is3,is4
prompt: .asciiz "Enter case input argument: "
currentvalue: .asciiz "Returned : i = " # Used for result output
space: .asciiz " \n" # Used for result output
.text
.globl main
main:
# prompt user for test value
li $v0,4 # syscall: print string
la $a0,prompt
syscall
# get user test value from stdin
li $v0,5 # syscall: read integer
syscall
move $a0,$v0 # move to argument for case
# LEARNER: testing use of the arg reg to hold input
###li $a0,0 # test value for case function
# LEARNER: set $ra and go to the case procedure (i.e. we are calling the
# case function)
jal case # calculate [whatever case calculates ;-)]
# print the prefix string
li $v0,4 # syscall: print string
# LEARNER: address of currentvalue into $a0
la $a0,currentvalue
syscall
# print the value the 'case' function returned
li $v0,1 # syscall: print integer
# LEARNER: move the contents of $t3 to $a0
move $a0,$t3 # get value to reg for syscall
syscall
li $v0,10 # syscall: exit program
syscall
# case -- does what?
#
# RETURNS: what does it return?
# t3 -- return value
# FIXME: to follow ABI, this should be v0 but okay for small internal
# functions like this one.
#
# arguments:
# a0 -- contains 'i' (FIXME: what is 'i'?)
#
# internal register usage (i.e. what registers get clobbered):
# t0 -- offset into jump table
# t1 -- address of jump table
# t2 -- internal jump address
case:
move $t3,$a0 # save 'i' to our work/return register
blt $a0,1,isD # jump to default if i < 1
bgt $a0,4,isD # jump to default if i > 4
# LEARNER: multiply 'i' by 2^2 (4) --> i <<= 2
sll $t0,$a0,2 # get the offset into the jump table
la $t1,jumptable # base address of the jump table
# LEARNER: $t1 = $t1(base address of table) + $t0('i' * 4)
add $t1,$t1,$t0 # add offset to base address
# LEARNER: $t2 is loaded with the address of label found in $t1
lw $t2,($t1) # jump_address = jumptable[i]
jr $t2 # jump to desired label
is1:
addi $t3,$t3,1 # i += 1
jr $ra # return to caller
is2:
addi $t3,$t3,2 # i += 2
jr $ra # return to caller
is3:
addi $t3,$t3,3 # i += 3
jr $ra # return to caller
is4:
addi $t3,$t3,4 # i += 4
jr $ra # return to caller
isD:
# BUG: this should be t3
###li $t0,0 # i = 0
li $t3,0 # i = 0
jr $ra # return to caller