Question

好吧，因为我是ASM的新手，我在创建加密邮件的程序时遇到了一些问题。此时我只是想输入一个字符，输出应该是字母表中的char + 5个位置。

因此程序应该读取ASCII中的字符并向其中添加5，然后打印该字母。

例：输入：A 输出：F

它应仅适用于大写字母，因此每个字符应为＆gt; = 65且<= 90。所以，如果我写'Z＆＃39;它应该从头开始字母并打印E＆＃39;。

到目前为止，我的代码看起来像这样：

    li $v0, 8               #read_string
    syscall                 #adresses char at $v0


    li $v0, 5               #char ASCII (I GUESS IT SHOULD)
    move $t0, $v0           #moves char(ASCII) to $t0
    li $s0, 90
    li $s1, 65
    bgt $t0, $s0, le_string # checks if char(ASCII) > 90
    ble $t0, $s1, le_string # checks if char(ASCII) < 65



    li $s0, 5               
    add $t1, $t1, $s0       #char=char+5
    move $t2, $t1           #moves encrypted char to $t2


    li $v0, 4               #print_string
    move $a0, $t2       
    syscall

LOG

25 / DEZ / 2015：输入：A 输出：A

如果有人能帮助我，我真的很感激！谢谢，

Rui（葡萄牙）

Answer 1

如果您使用的是MIPS，则代码应如下所示：

    .text
main:
    li $v0, 8                 # read_string
    la $a0, textbuf           # adresses of char at $a0
    li $a1, 2                 # length to read at $a1
    syscall

    la $t0, textbuf
    lb $v0, 0($t0)            # char ASCII (I GUESS IT SHOULD)
    move $t0, $v0             # moves char(ASCII) to $t0
    li $s0, 90
    li $s1, 65
    bgt $t0, $s0, le_string   # checks if char(ASCII) > 90
    nop                       # avoid instruction after branch begin executed even if jump is taken
    blt $t0, $s1, le_string   # checks if char(ASCII) < 65
    nop                       # avoid instruction after branch begin executed even if jump is taken

    li $s2, 5
    li $s3, 26
    add $t0, $t0, $s2         # char=char+5
    ble $t0, $s0, nowrap_char # checks if encrypted char <= 90
    nop                       # avoid instruction after branch begin executed even if jump is taken
    sub $t0, $t0, $s3         # wrap the char
nowrap_char:
le_string:
    move $t2, $t0             # moves encrypted char to $t2

    la $t0, textbuf
    li $v0, 4                 # print_string
    sb $t2, 0($t0)            # put the encrypted char
    la $a0, textbuf
    syscall

    li $v0, 10                # exit
    syscall

    .data
textbuf:
    .space 2

＆＃34;加密＆＃34;大会中的消息

1 个答案: