assembly - 反转 MIPS 汇编中数字的位

Question

我正在学习 MIPS 汇编语言，我被要求编写一个执行以下操作的程序:

1. 接受一个整数作为用户输入
2. 将该整数打印为带符号的二进制数
3. 反转该整数中的位
4. 打印结果反转数(也是二进制)

除了我显然需要使用移位操作和两个移位值之间的逻辑比较这一事实之外，我还很迷茫。我的想法是编写一个循环，将输入整数的最低有效位放入临时寄存器，将该值左移 1，将输入整数右移 1，然后重复该过程，直到输入整数等于 0 .我只是不确定如何“取”一个数字...有人可以帮我吗？

Answer 1

我解决了:

.data
    int: .word 0

    pleaseEnter: .asciiz "Please enter an integer: "
    yourIntBin: .asciiz "Your integer in binary code is:        "
    yourIntBinRev: .asciiz "And now in reverse:         "
    nl: .asciiz "\n"
.text
    main:
    # ask the user for an integer
    li $v0, 4
    la $a0, pleaseEnter
    syscall

    # get the user's integer, store in int
    li $v0, 5
    syscall
    sw $v0, int

    # set up temporary reg's
    li $t7, 31 # shamt is in t7

    j loop1

    loop1:
        lw $t0, int # load int to t0
        srlv $t1, $t0, $t7 # shift by counter
        bnez $t1, exit1 # if the bit is 1, exit the loop
        beqz $t7, exit1 # if the counter has reached 0 with no 1s in the int word, exit the loop
        addi $t7, $t7, -1
        j loop1

    exit1: # now t7 is the number of bits that come before the msb

    move $t4, $t7 # save that number for later

    # tell the user what their number in binary is:
    li $v0, 4
    la $a0, nl
    syscall
    li $v0, 4
    la $a0, yourIntBin
    syscall
    li $v0, 1

    # prepare temp reg's
    li $t6, 32 # set t6 to the maximum number of bits in a word
    sub $t7, $t6, $t7 # make t7 the number of bits that come after the msb (from the right)

    j loop2

    loop2:
        # print the current bit
        move $a0, $t1
        syscall

        beq $t7, $t6, exit2 # if t7 is now 32, leave the loop

        lw $t0, int # load the integer
        sllv $t0, $t0, $t7 # shift left by counter
        srl $t1, $t0, 31 # isolate the bit
        addi $t7, $t7, 1 # increment shamt
        j loop2

    exit2:

    li $v0, 4
    la $a0, nl
    syscall
    la $a0, yourIntBinRev
    syscall

    # now it's time to print the binary code in reverse
    addi $t6, $t6, -1 # get t6 to be 31
    move $t7, $t4 # put t7's value after the first loop back in t7
    li $t4, 0 # clear out t4 to save register space
    li $v0, 1 # prepare to print integers ("bits")
    li $t5, 0 # ensure t5 is 0
    lw $t0, int # get the input integer to t0

    loop3:
        sub $t4, $t6, $t5 # get t4 to the desired shamt
        sllv $t1, $t0, $t4 # shift left to eliminate unwanted bits from the left
        srl $a0, $t1, 31 # same, but now for the 31 bits to the right of my desired bit
        syscall # print that bit
        beq $t5, $t7, exit3 # if the reversal is complete, leave the loop
        addi $t4, $t4, 1 # add 1 to the shamt
        addi $t5, $t5, 1 # add 1 to the counter
        j loop3
    exit3:

    li $v0, 4
    la $a0, nl
    syscall

# end of execution
li $v0, 10
syscall