无法搞清楚为什么我的code,用于检查数量是palindrom将无法正常工作。咨询AP preciated我的、无法正常、数量、工作

由网友(怪咖小青年)分享简介:我目前对Java,所以我通常的语言和编程的理解是升技弱。我的code是在这里:I'm currently new to Java, so my understanding of the language and programming in general is abit weak. My code is here:...

我目前对Java,所以我通常的语言和编程的理解是升技弱。我的code是在这里:

I'm currently new to Java, so my understanding of the language and programming in general is abit weak. My code is here:

import java.util.Scanner;
public class task2 {
public static void main(String args[])  {
    System.out.print("Input a 3 digit int");

    Scanner scan = new Scanner(System.in);
int x = scan.nextInt();


int isPalindrome = 0; 

while (x != 0)
{
    isPalindrome = isPalindrome*10 + x % 10;
    x /= 10;
}


{


if (x == isPalindrome){
    System.out.print ("Yes, this is a palindrome!");
}
else {
    System.out.print("No, try again");
}}}}

在code只能识别回文如果输入的数字是零。我无法理解为什么。感谢您的咨询!

The code will only recognize a palindrome if the numbers entered are zeroes. I'm having trouble understanding why. Thank you for the advice!

推荐答案

这是因为x的值是得到改变finally.Which是不是在程序结束时,原来的号码。 所以要略低于x像另一个变量: INT Y = X; 并且,同时使用的如果条件中使用的y此值作比较,而不是使用X在末端。它将运行完美。

This is because the value of x is getting changed finally.Which is not the original number at the end of the program. SO take another variable just below x like: int y = x; And at the end while using "if" condition use this value of y for comparison rather than using x. It will run perfectly.

INT X = scan.nextInt();

int x = scan.nextInt();

INT Y = X;

int y=x;

如果(Y == isPalindrome)添加新的变量是这样的。

if (y == isPalindrome) Add new variable like this.

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