在函数中将 Scanner 对象作为参数传递的基本语法

问题描述

这是我写的非常基本的内容:

Here is what I wrote which is pretty basic :

import java.util.Scanner;

public class Projet {

    /**
     * @param args
     * @param Scanner
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println("Enter a digit");
        Scanner in = new Scanner(System.in);
        getChoice(Scanner);
        in.close();
    }

    public static int getChoice(Scanner n){
        n = in.nextInt();
        return n;
    }
}

这里似乎有什么问题?我早些时候让它工作，我必须将扫描仪类型和参数名称作为参数传递给函数......然后简单地使用在主函数中调用该函数>扫描仪类型和参数作为函数的参数?

What seems to be wrong here ? I had it working earlier, I had to pass the Scanner type and argument name as a parameter to the function... and simply call that function in the main using Scanner type and argument as an argument to the function ?

-----编辑-----

下面需要它的新代码:

import java.util.Scanner;

public class Projet {

    /**
     * @param args
     * @param Scanner
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println("Enter a digit");
        Scanner in = new Scanner(System.in);
        System.out.println(getChoice(in));
        in.close();
    }

    public static int getChoice(Scanner in){
        return in.nextInt();
    }
}

@rgettman 谢谢！

@rgettman Thanks !

推荐答案

调用方法时需要传入实际的变量名in，而不是类名Scanner>.

You need to pass the actual variable name in when you call the method, not the class name Scanner.

getChoice(in);

代替

getChoice(Scanner);

顺便说一下，您的 getChoice 方法不会如所示进行编译.只需返回扫描器返回的内容，即 int，正如您声明 getChoice 返回一个 int:

Incidentally, your getChoice method won't compile as shown. Just return what the scanner returns, which is an int, as you declared getChoice to return an int:

public static int getChoice(Scanner n){
    return n.nextInt();
}

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