Question

这个程序应该是一个猜谜游戏，你在a和j之间输入一个字母，并尝试猜测随机选择的答案。我一直收到有关if语句的错误消息。对于那些＆lt;或＆gt;，它表示“错误：二元运算符的错误操作数类型”。对于带有==或！=的那些，它表示错误：无法比较的类型：char和Random。我该怎么做才能解决这个问题？

import java.util.Scanner;
import java.lang.Math;
import java.util.Random;

class Guess{

    public static void main( String[] args ){

    Scanner sc = new Scanner(System.in);

    System.out.println("I'm thinking of a letter in the range a to j. You have three guesses. ");

    System.out.print("Enter your first guess: ");

    Random r = new Random();
    char i = (char)(r.nextInt(10) + 'a');

    if(i>r){
        System.out.println("Your guess is too high. ");

    } else if(i<r){
        System.out.println("Your guess is too low. ");

    } else if(i==r){
        System.out.println("You win! ");

    }

    System.out.print("Enter your second guess: ");

    i = (char)(r.nextInt(10) + 'a');

    if(i>r){
        System.out.println("Your guess is too high. ");

    } else if(i<r){
        System.out.println("Your guess is too low. ");

    } else if(i==r){
        System.out.println("You win! ");

        }

    System.out.print("Enter your third guess: ");

    i = (char)(r.nextInt(10) + 'a');

    if(i>r){
        System.out.println("Your guess is too high. ");

    } else if(i<r){
        System.out.println("Your guess is too low. ");

    } else if(i==r){
        System.out.println("You win! ");

        }

    if(i!=r){
        System.out.println("You lose. The letter was " +r+ ".");
    }

    }
}

Answer 1

char是基本类型，Random是引用类型。您无法使用＆lt ;,＆gt;，==或！=运算符自由比较它们。

从我在您的代码中看到的内容，您实际上从未获得用户的输入。您应该使用扫描仪从标准输入读取并将读取值与i进行比较。

“二元运算符的错误操作数类型”和“无法比较的类型：char和Random”错误

1 个答案: