mips汇编语言中的位屏蔽/移位？

该计划的要求与我的解决方案一起发布在下面的代码中。有人能告诉我我做错了什么，并告诉我如何修复它吗？我每次输入的结果都是0。在大多数情况下，我相信我是在掩盖错误，而不是错误地移位

    ##########################################################################

############################
# prompt user to enter an integer, read the integer, and display a 0 if the bits at the
# 16 and 256 place value positions of the integer are both 1 and display a 1 otherwise
############################ data segment ################################
            .data
outputLegend1:      .asciiz "0 = both bits at 16 & 256 place value positions are 1\n"
outputLegend2:      .asciiz "1 = bits at 16 & 256 place value positions NOT both 1\n\n"
inputPrompt:        .asciiz "Enter integer: "
outputLabel:        .asciiz "Integer entered is of type "
############################ code segment ################################
            .text
            .globl main
main:
            li $v0, 4
            la $a0, outputLegend1        
            syscall         # print output legend part 1
            la $a0, outputLegend2        
            syscall         # print output legend part 2
            la $a0, inputPrompt        
            syscall         # print integer prompt
            li $v0, 5
            syscall         # read integer
            move $t0, $v0       # save integer read in $t0
            li $v0, 4
            la $a0, outputLabel        
            syscall         # print output label

            li $v0, 1

            ##########################################################
            # Insert NO MORE THAN 6 lines of code that involve ONLY 
            #   bit manipulating instructions (ANDing, ORing, XORing,
            #   NORing and shifting - only whatever that are needed)
            # so that the program will work just like the sample runs 
            # shown at the bottom (some blank lines edited out).








#my solution


            sll $t1,$t1,4
            andi $t2,$t1,1
            sll $t2,$t2,4
            andi $t3,$t2,1
            xor $a0, $t2,$t3






            syscall         # display desired output

                    ##########################################################

            li $v0, 10      # exit gracefully
                    syscall

########################## sample test runs ##############################
# 0 = both bits at 16 & 256 place value positions are 1
# 1 = bits at 16 & 256 place value positions NOT both 1
# 
# Enter integer: 0
# Integer entered is of type 1
# -- program is finished running --
# 
# Reset: reset completed.
# 
# 0 = both bits at 16 & 256 place value positions are 1
# 1 = bits at 16 & 256 place value positions NOT both 1
# 
# Enter integer: 16
# Integer entered is of type 1
# -- program is finished running --
# 
# Reset: reset completed.
# 
# 0 = both bits at 16 & 256 place value positions are 1
# 1 = bits at 16 & 256 place value positions NOT both 1
# 
# Enter integer: 256
# Integer entered is of type 1
# -- program is finished running --
# 
# Reset: reset completed.
# 
# 0 = both bits at 16 & 256 place value positions are 1
# 1 = bits at 16 & 256 place value positions NOT both 1
# 
# Enter integer: 272
# Integer entered is of type 0
# -- program is finished running --
# 
# Reset: reset completed.
# 
# 0 = both bits at 16 & 256 place value positions are 1
# 1 = bits at 16 & 256 place value positions NOT both 1
# 
# Enter integer: 12349876
# Integer entered is of type 0
# -- program is finished running --
# 
# Reset: reset completed.
# 
# 0 = both bits at 16 & 256 place value positions are 1
# 1 = bits at 16 & 256 place value positions NOT both 1
# 
# Enter integer: 12346789
# Integer entered is of type 1
# -- program is finished running --
# 
######################## end sample test runs ############################

你的想法是对的（但是有很多拼写错误），所以继续吧

---

第一次移动时，

$t1

没有什么有趣的。我认为您最初希望将

$t0

转换为

$t1

，因此您仍然拥有用户输入的原始值，以便以后进行操作

---

注意你的注册用法：

• 您需要源代码寄存器，这些寄存器在您的程序中已被锁定
• 如果以后需要一些原始/源代码值（不要删除仍然需要的值），则需要为目标使用新的/不同的寄存器
• 当您需要旧的/原始的值时，您不能简单地寻找新的寄存器

使用单步验证是否在每个指令之后获得了预期的中间值。如果没有，则检查指令是否源于正确的寄存器，或者，源寄存器可能已被先前的指令删除

---

而且，当你想右移时，你正在左移

---

您只是在猜测何时使用哪个寄存器。因此，这里有一种方法可以帮助您摆脱由于不熟悉的组装而带来的混乱

试着用C编写你的代码，使用——对于这个作业，它不会花费太多，大约5行！但这将帮助您了解何时使用哪个变量。（确保您的C三地址代码正常工作-在某个地方测试并调试它，如有必要，请使用联机C编译器。）

接下来，在编写任何指令之前，将所有这些C（TAC）变量分配（创建映射）到MIPS寄存器

最后，使用变量到寄存器的映射，编写尊重/模拟这三个地址代码的汇编指令