所以,我不知道装配是如何工作的,也不知道我在做什么。我以为我做了,但当然我错了。所以这就是我的问题 - 我甚至不知道如何让用户输入一个整数,这样我就可以将它存储在内存中。我也不知道我的变量是否一致,因为我甚至不理解"对齐"真的是。下面是我的汇编代码,以及说明我要为代码做什么的评论。请帮忙
.data
# variables here
intPrompt: .asciiz "\nPlease enter an integer.\n"
stringPrompt: .asciiz "\nPlease enter a string that is less than 36 (35 or less) characters long.\n"
charPrompt: .asciiz "\nPlease enter a single character.\n"
int: .space 4
string: .space 36
char: .byte 1
.text
.globl main
main:
# print the first prompt
li $v0, 4
la $a0, intPrompt
syscall
# allow user to enter an integer
li $v0, 5
syscall
# store the input in `int`
# don't really know what to do right here, I want to save the user inputed integer into 'int' variable
sw $v0, int
syscall
答案 0 :(得分:2)
你应该改变" int"的类型。变量从.space到.word
最后看起来应该是这样的:
.data
# variables here
intPrompt: .asciiz "\nPlease enter an integer.\n"
stringPrompt: .asciiz "\nPlease enter a string that is less than 36 (35 or less) characters long.\n"
charPrompt: .asciiz "\nPlease enter a single character.\n"
int: .word
string: .space 36
char: .byte 1
main:
li $v0, 4 #you say to program, that you're going to output string which will be in the $a0 register
la $a0, intPrompt #here you load your string from intPromt var. to $a0
syscall #this command just executes everything that you have written before >> it prints string, which is in $a0
li $v0, 5 #this command says: "Hey, read an integer from console and put it in $v0!"
syscall #this command executes all previous commands ( li $v0, 5 )
sw $v0, int #sw -> store word, here you move value from $v0 to "int" variable
syscall #executes (sw $v0, int), here you have your input number in "int" variable