Question

我的代码是一个非常基本的程序，而不是我输入System.out.println时，我得到了错误：＆＃34;对println的引用含糊不清＆＃34;并且随机出现它用来工作并且没有错误，但现在它确实发生了这个错误，所以我怎么解决这个问题，而不是你。

    package blue.light;
    //name is: slash
    import java.util.*;

    public class BlueLight {

        public static void main(String[] args) throws InterruptedException {
    Scanner scan = new Scanner(System.in);

    String name;
    int age;
    String a;
    String yes;
    String no;
    double num1;
    double num2;
    double ans;





    System.out.println("hello, my name is Slash");
    System.out.println("What is your name?");
    name = scan.nextLine();
    System.out.println("Hello " + name);
    Thread.sleep(1000);
    System.out.println("How old are you?");
    age = scan.nextInt();
    if(age < 18)
            {
                System.out.println("ur young af");
            }
    else if (age == 21)
    {
        System.out.println("Shots, shots, shots");

    }
    else if (age == 69)
    {
        System.out.println("ohhh yea");
    }
    System.out.println("you to calculate any thing?");
    a = scan.nextLine();
    if(a==("yes")){
        System.out.println("What would u like to do?(multiply,or add)");     
        a = scan.nextLine();
                    if(a == "add"){
            System.out.println("Enter number 1");
            num1 = scan.nextDouble();
            System.out.println("Enter number 2");
            num2 = scan.nextDouble();
                    ans = num1+num2;

        }


    }

}
    }

Answer 1

首先，nextInt()在年龄之后不会消耗新行（因此您的nextLine为空）。您将Object值与.equals()调用进行比较，因为==测试引用标识（并使用String我使用String.equalsIgnoreCase(String)）。最后，我将在Java中声明具有最小可见性的变量;尽可能接近他们的第一次使用。像，

public static void main(String[] args) throws InterruptedException {
    Scanner scan = new Scanner(System.in);

    System.out.println("hello, my name is Slash");
    System.out.println("What is your name?");
    String name = scan.nextLine();
    System.out.println("Hello " + name);
    Thread.sleep(1000);
    System.out.println("How old are you?");
    int age = scan.nextInt();
    if (age < 18) {
        System.out.println("ur young af");
    } else if (age == 21) {
        System.out.println("Shots, shots, shots");
    } else if (age == 69) {
        System.out.println("ohhh yea");
    }
    System.out.println("you to calculate any thing?");
    scan.nextLine(); // <-- nextInt leaves a trailing new line.
    String a = scan.nextLine();
    if (a.equalsIgnoreCase("yes")) { // <-- compare string for equality in
                                        // Java
        System.out.println("What would u like to do?(multiply,or add)");
        a = scan.nextLine();
        if (a.equalsIgnoreCase("add")) {// <-- compare string for equality
                                        // in Java
            System.out.println("Enter number 1");
            double num1 = scan.nextDouble();
            System.out.println("Enter number 2");
            double num2 = scan.nextDouble();
            double ans = num1 + num2;
            System.out.println(ans); // <-- display answer
        }
    }
}

对println的引用是不明确的错误

1 个答案: