MIPS 查找字符串结尾
MIPS find end of string
我有以下 mips 代码(运行 它在 QTSPIM 中),它应该计算字符串中的字符数并打印它们。
其背后的逻辑非常简单,但并没有按预期工作。一切顺利,直到它到达字符串的末尾,然后它继续计数,即使我将每个元素与 $zero 进行比较以找到字符串的末尾 (\0)。
是不是我退出循环的条件有问题,还是my_string最后没有包含\0所以不会退出?
.data
endl: .asciiz "\n"
my_string: .asciiz "thisisastring"
star: .asciiz "*"
str_end: .word 0
space: .asciiz " "
.text
.globl main
main:
la $a0, my_string
li $v0, 4
syscall
la $a0, endl
li $v0, 4
syscall
la $t0, my_string # load mystring to $t0
li $t1, 0 # make $t1 = 0, character counter
lb $t2, ($t0) # make $t2 point to the first character of "my_string"
li $t3, 1 # $t3 is the ++ register to go to the next character
li $t4, 0 # character counter
la $t5, str_end
cont:
beqz $t0, print # if [=10=] is found print and exit
addi $t4, $t4, 1 # increase the counter
lbu $a0, ($t0) # print current character
li $v0, 11
syscall
addi $t0, $t0, 1 # go to next char
#move $t2, $t0
j cont
print:
move $a0, $t4
li $v0, 1
syscall
j exit
exit:
li $v0, 10
syscall
问题出在指令的顺序,代码的逻辑上。
这是没有冗余代码的更正版本:
.data
endl: .asciiz "\n"
my_string: .asciiz "thisisastring"
str_end: .word 0
.text
.globl main
main:
la $a0, my_string
li $v0, 4 # print the string
syscall
la $a0, endl # print endl
li $v0, 4
syscall
la $t0, my_string # load mystring to $t0
li $t1, 0 # make $t1 = 0, character counter
lb $t2, 0($t0) # make $t2 point to the first character of "my_string"
la $t5, str_end
cont:
lb $a0, 0($t0) # print current character
beqz $a0, print # if [=10=] is found print and exit
addi $t1, $t1, 1 # increase the counter
addi $t0, $t0, 1 # go to next char
li $v0, 11
syscall
j cont
print:
move $a0, $t1
li $v0, 1
syscall
j exit
exit:
li $v0, 10
syscall
我有以下 mips 代码(运行 它在 QTSPIM 中),它应该计算字符串中的字符数并打印它们。
其背后的逻辑非常简单,但并没有按预期工作。一切顺利,直到它到达字符串的末尾,然后它继续计数,即使我将每个元素与 $zero 进行比较以找到字符串的末尾 (\0)。
是不是我退出循环的条件有问题,还是my_string最后没有包含\0所以不会退出?
.data
endl: .asciiz "\n"
my_string: .asciiz "thisisastring"
star: .asciiz "*"
str_end: .word 0
space: .asciiz " "
.text
.globl main
main:
la $a0, my_string
li $v0, 4
syscall
la $a0, endl
li $v0, 4
syscall
la $t0, my_string # load mystring to $t0
li $t1, 0 # make $t1 = 0, character counter
lb $t2, ($t0) # make $t2 point to the first character of "my_string"
li $t3, 1 # $t3 is the ++ register to go to the next character
li $t4, 0 # character counter
la $t5, str_end
cont:
beqz $t0, print # if [=10=] is found print and exit
addi $t4, $t4, 1 # increase the counter
lbu $a0, ($t0) # print current character
li $v0, 11
syscall
addi $t0, $t0, 1 # go to next char
#move $t2, $t0
j cont
print:
move $a0, $t4
li $v0, 1
syscall
j exit
exit:
li $v0, 10
syscall
问题出在指令的顺序,代码的逻辑上。
这是没有冗余代码的更正版本:
.data
endl: .asciiz "\n"
my_string: .asciiz "thisisastring"
str_end: .word 0
.text
.globl main
main:
la $a0, my_string
li $v0, 4 # print the string
syscall
la $a0, endl # print endl
li $v0, 4
syscall
la $t0, my_string # load mystring to $t0
li $t1, 0 # make $t1 = 0, character counter
lb $t2, 0($t0) # make $t2 point to the first character of "my_string"
la $t5, str_end
cont:
lb $a0, 0($t0) # print current character
beqz $a0, print # if [=10=] is found print and exit
addi $t1, $t1, 1 # increase the counter
addi $t0, $t0, 1 # go to next char
li $v0, 11
syscall
j cont
print:
move $a0, $t1
li $v0, 1
syscall
j exit
exit:
li $v0, 10
syscall