Question

为什么变量a1没有初始化？我将其定义为if条件，但不是初始条件。

import java.util.InputMismatchException;
import java.util.Scanner;

public class Input {
    static Scanner input; 

    public Input() {
        input = new Scanner(System.in);
        System.out.println("Welcome ");
    }

    public static Riazy GetInfo() throws InputMismatchException {
        System.out.println("Enter the first Number ");

        if(input.hasNextDouble()==false){
            throw new InputMismatchException("Erorr...");
        } else {
            double a1 = input.nextDouble();
        }
        System.out.println("Enter the second Number ");

        double a2 = input.nextDouble();
        boolean a4 = input.hasNextDouble();

        if(a4==false){
            throw new InputMismatchException("Erorr...");
        }
        Riazy riazy = new Riazy(a1 , a2);

        return riazy;
    }
}

Answer 1

像 Mike'Pomax'Kamermans 所说：“您没有正确设置变量的范围” 。

不会编译

public static Riazy GetInfo() throws InputMismatchException {
    System.out.println("Enter the first Number ");
    if (!input.hasNextDouble()) {
        throw new InputMismatchException("Erorr...");
    } 
    else {
        double a1 = input.nextDouble();
        // a1 is only accessible inside this block
    }
    System.out.println("Enter the second Number");
    double a2 = input.nextDouble();
    boolean a4 = input.hasNextDouble();
    if (!a4) {
        throw new InputMismatchException("Erorr...");
    }
    /*
     * Won't compile as a1 has been defined inside the if 
     * else block and is not visible here 
     */
    return new Riazy(a1, a2);
}

编译

public static Riazy GetInfo() throws InputMismatchException {
    System.out.println("Enter the first Number ");
    double a1 = null; // a1 is visible in this scope
    if (!input.hasNextDouble()) {
        throw new InputMismatchException("Erorr...");
    } 
    else {
        a1 = input.nextDouble();
    }
    System.out.println("Enter the second Number");
    double a2 = input.nextDouble();
    boolean a4 = input.hasNextDouble();
    if (!a4) {
        throw new InputMismatchException("Erorr...");
    }
    return new Riazy(a1, a2);
}

我建议您阅读：Java doc for Scope of variables

为什么a1可变剂量不是初始的？

1 个答案: