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我需要帮助来编写一个生成回文的程序。我设法使字符串反向,但我无法将原始字符串和反向字符串组合在一起。当我写 abc 时,我需要得到 abccba,或者当我写 hello 时,我需要得到 helloolleh。现在我只得到 cba 或 olleh。有人可以帮我解决这个问题吗?
.data
msg1: .asciiz "Enter the length of your input: "
msg2: .asciiz "Enter your input: "
msg3: .asciiz "Output of the program is: "
output: .space 256 # will store the output palindrome
.text
la $a0,msg1
li $v0,4
syscall
li $v0,5
syscall
move $s1,$v0 # $s1 has the length of the string
la $a0,msg2
li $v0,4
syscall
li $a1,1024
li $v0,8
syscall
move $s0,$a0 # $s0 has the starting address of the string
#
# YOUR CODE GOES HERE(You can use additional labels at the end)
#
move $a3,$s0
move $a1,$s1
add $a3,$a3,$a1
la $a2, output($zero)
jal reverse
la $a0,msg3
li $v0,4
syscall
la $a0,output($zero)
li $v0,4
syscall
li $v0,10
syscall
reverse:
# $a0 - address of string to reverse
# a1 - length of the string
# a2 - address of string where to store the reverse
addi $sp, $sp, -4
sw $ra, 0($sp)
bltz $a1, reverse_end
lb $t0, 0($a3)
subi $a1, $a1, 1
subi $a3, $a3, 1
sb $t0, 0($a2)
addi $a2, $a2, 1
jal reverse
reverse_end:
lw $ra, 0($sp)
addi $sp, $sp, 4
jr $ra
#include <iostream>
#include <string>
using namespace std;
string palindrom(string input, int rem_length)
{
if(rem_length!=0)
{
input=input.substr(0,1)+palindrom(input.substr(1,input.length()-1), \\
rem_length-1)+input.substr(0,1);
}
return input;
}
int main()
{
string input;
cin >> input;
input = palindrom(input, input.length());
cout << input<< endl;
system("pause");
return 0;
}
最佳答案
看来您只需要先将输入字符串复制到 output
字符串,当你通过 output
数组到 reverse
函数,您需要将指针前移输入字符串长度(传递 output + strlen(input)
),以便将反转的字符串写在它旁边。
另外我认为您应该为输入使用缓冲区。还要照顾好'\n'
char 在输入字符串的末尾,因为它将在那里。
考虑到这些,我将您的代码更改为此
.data
msg1: .asciiz "Enter the length of your input: "
msg2: .asciiz "Enter your input: "
msg3: .asciiz "Output of the program is: "
input: .space 128
output: .space 256 # will store the output palindrome
.text
la $a0,msg1
li $v0,4
syscall
li $v0,5
syscall
move $s1,$v0 # $s1 has the length of the string
la $a0,msg2
li $v0,4
syscall
move $a1, $1
la $a0, input
li $v0,8
syscall
move $s0,$a0 # $s0 has the starting address of the string
#
# YOUR CODE GOES HERE(You can use additional labels at the end)
#
move $a3,$s0 # a3 = (char *)input
move $a1,$s1 # a1 = strlen(input)
# replacing the read in new line char with 0
la $t4, input
add $t4, $t4, $a1
sb $zero, ($t4)
# copy the string to the resulting string
move $t0, $s1 # i = strlen(input)
move $t1, $s0
la $t3, output
copy_loop:
beq $t0, $zero, copy_end # i != 0
lb $t2, ($t1)
sb $t2, ($t3) # *output = *input
addi $t1, $t1, 1 # advancing the input ptr
addi $t3, $t3, 1 # advancing the output ptr
subi $t0, $t0, 1 # i--
b copy_loop
copy_end:
add $a3,$a3,$a1 # a3 = &input[strlen(input)]
subi $a3,$a3,1 # subtracting one since the last input char is at input[strlen(input)-1]
la $a2, output
add $a2, $a2, $a1 # passing the address of output advanced by the input length
jal reverse
la $a0,msg3
li $v0,4
syscall
la $a0,output($zero)
li $v0,4
syscall
li $v0,10
syscall
reverse:
# $a0 - address of string to reverse
# a1 - length of the string
# a2 - address of string where to store the reverse
addi $sp, $sp, -4
sw $ra, 0($sp)
bltz $a1, reverse_end
lb $t0, 0($a3)
subi $a1, $a1, 1
subi $a3, $a3, 1
sb $t0, 0($a2)
addi $a2, $a2, 1
jal reverse
reverse_end:
lw $ra, 0($sp)
addi $sp, $sp, 4
jr $ra
move $a1, $1
la $a0, output # !! using the output buffer for the input
li $v0,8
syscall
move $s0,$a0 # $s0 has the starting address of the string
#
# YOUR CODE GOES HERE(You can use additional labels at the end)
#
move $a3,$s0 # a3 = (char *)input
move $a1,$s1 # a1 = strlen(input)
# the whole copy thing deleted and don't need to be bothered by the new
# lines, it will be overwritten with the reversed string anyway
add $a3,$a3,$a1 # a3 = &input[strlen(input)]
subi $a3,$a3,1 # subtracting one since the last input char is at input[strlen(input)-1]
la $a2, output
add $a2, $a2, $a1 # passing the address of output advanced by the input length
jal reverse
关于assembly - MIPS中的回文生成,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60441799/
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