Question

我不知道为什么我会收到这个错误。我认为一般的代码是可以的，虽然我确信有一个更短的方法，然后使用所有其他ifs。问题是它说不兼容的类型，我真的只是失去了如何解决这个问题。任何帮助，将不胜感激。

import java.util.Scanner;
public class MissionImpossible
{
    public static void main(String [] args){
        String lineOne,  R2D2 = "";

       Scanner in = new Scanner(System.in);
       System.out.println("Please enter a word so I can see how many vowels it has.");

       int count = 0;

       lineOne = in.nextLine();

       int word = lineOne.length();

       for (int i = word -1; i>= 0; i--) 
        {
            R2D2= lineOne.charAt(i);

            if (R2D2== 'a'|| R2D2=='A')
            count++;
            else if (R2D2=='e'||R2D2=='E')
            count++;
            else if (R2D2=='o'|| R2D2=='O')
            count++;
            else if (R2D2=='u'||R2D2=='U')
            count++;
            else if (R2D2=='y'||R2D2=='Y')
            count++;

    }
    System.out.println(count);
          }
           }

Answer 1

char不是String。将R2D2声明为char

char R2D2 = '';

要检查元音，请制作如下方法，并在for循环和count元音发生时重复使用此方法：

static boolean isVowel(char ch) {
    ch = Character.toLowerCase(ch);
    if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') {
        return true;
    }
    return false;
 }

Answer 2

    String lineOne,  R2D2 = "";

R2D2是一个字符串，您正在与字符if (R2D2== 'a'|| R2D2=='A')

进行比较

试试这个

for（int i = word -1; i＆gt; = 0; i--） { R2D2 = lineOne.charAt（i）;

        if (R2D2=="a"|| R2D2=="A")
        count++;
        else if (R2D2=="e"||R2D2=="E")
        count++;
        else if (R2D2=="o"|| R2D2=="O")
        count++;
        else if (R2D2=="u"||R2D2=="U")
        count++;
        else if (R2D2=="y"||R2D2=="Y")
        count++;

}

一方面不是，您应该使用.equals()代替==

进行比较

Answer 3

如果你想修复long if链，你可以这样做：

if (anyOf(R2D2, "AaEeIiOoUuYy".toCharArray())

后来有：

private static boolean anyOf(char in, char[] items) {
    for (int i = 0; i < items.length; i++) {
        if (in == items[i])
            return true;
    }
    return false;
}

Answer 4

R2D2的类型为String，而单个引号围绕一个字符（如'a'中所示）使其成为字符文字。字符串不能分配或与字符比较。这就是原因。

代码中存在两种问题。

1. `R2D2= lineOne.charAt(i); // a character being assigned to a String variable`

解决方案：R2D2 = Character.toString(lineOne.charAt(i));

2. `if (R2D2== 'a'|| R2D2=='A')` //strings being compared with char literals.

解决方案：if (R2D2.equals("a")|| R2D2.equals("A"))

仅供参考，为了改善维护，您也可以这样做。

if(R2D2.equalsIgnoreCase('a'))

不兼容的类型字符串和字符

4 个答案: