比较 MIPS 中的两个字节 [英] Comparing two bytes in MIPS

问题描述

我正在尝试制作一个简单的 MIPS 程序.我想检查用户输入的字符是否为空格.我在 .data 部分定义了一个变量 space.然后我接受用户输入，并使用 jal 调用我的函数来检查这个.我将字节 space 加载到 $a0 中，然后想检查 $t0 和 $a0 是否相同.如果它们不是 $t1 应该设置为 1 然后我使用 branch not equal to branch 到我的其他函数，该函数将响应打印给用户.但是，无论我输入什么字符，它总是分支.我不确定我做错了什么.

I'm trying to make a simple MIPS program work. I want to check if the character inputted by the user is a space. I have defined a variable space in my .data section. I then take the user input, and use jal to call my function which should check this. I load the byte space into $a0 and then want to check whether $t0 and $a0 are the same. If they aren't $t1 should be set to 1 and then I used branch not equal to branch to my other function which prints out the response to the user. However, no matter what character I enter it always branches. I'm not sure what I'm doing wrong.

space: .byte ' '

main:
    #Getting user input 
    li $v0,8
    la $a0,str1
    li $a1, 20
    syscall 
    jal is_space

    #Indicate the end of main function
    li $v0,10
    syscall
is_space: 
    add $t0, $a0, $zero
    lb $a0, space
    sltu $t1, $a0, $t0
    bne $t1, $zero, spaceinput

推荐答案

您甚至忘记访问已读取的字符.它位于 str1 的缓冲区中.另外，如果您想检查相等性，为什么要使用 sltu?将空间存储在内存中也是一种浪费.

You forgot to even access the character that was read. It's placed in a buffer at str1. Also, if you want to check for equality, why are you using sltu? Storing the space in memory is a waste, too.

main:
    #Getting user input 
    li $v0,8
    la $a0,str1
    li $a1, 20
    syscall 
    lb $a0, ($a0) # fetch first character entered
    jal is_space

    #Indicate the end of main function
    li $v0,10
    syscall
is_space: 
    li $t0, ' '
    beq $a0, $t0, spaceinput

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