Java中BufferedReader与Scanner读入的区别详解

BufferedReader br = new BufferedReader (newInputStreamReader(System.in));

package com.zxt.base; 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
 
public classSystemInTest { 
    private static Scanner sc; 
    public static void main(String[] args) {
        // 使用Scanner输入
        sc = new Scanner(System.in);
        int num1 = sc.nextInt();
        int num2 = sc.nextInt();
        System.out.println("num1 + num2 = "+ (num1+ num2));
 
        // 使用BufferReader输入
        InputStreamReaderisr = newInputStreamReader(System.in);
        BufferedReaderbr = new BufferedReader(isr);
        try {
            int num3 = Integer.parseInt(br.readLine());
            int num4 = Integer.parseInt(br.readLine());
            System.out.println("num3 + num4 = "+ (num3+ num4));
 
        }catch(NumberFormatException | IOException e) {
            e.printStackTrace();
        }       
       
        // 使用Scanner输入会遇到的问题
        System.out.println();
        System.out.print("Enter an Integer:");
        int intValue = sc.nextInt();
        System.out.print("Enter a String:");
        StringstrValue = sc.nextLine();
        System.out.printf("You have entered intValue is " + intValue+ " and strValue is " + strValue);
       
        // 问题原因：由于nextXXX()方法忽略换行符，但是nextLine()并不忽略它。所以如果我们在nextXXX()方法后面使用nextLine()将会出现问题
        // 解决方案是：在使用nextXXX()方法后，在使用nextLine()读取下一行数据前，多使用一个nextLine()用来消耗换行符
        // int intValue = sc.nextInt();
        // sc.nextLine();
        // String strValue = sc.nextLine();
       
        // 或者使用BufferReader不会出现该问题
        System.out.println();
        try {
            System.out.print("Enter an Integer:");
            int intValue1 = Integer.parseInt(br.readLine());
            System.out.print("Enter a String:");
            StringstrValue1 = br.readLine();
            System.out.printf("You have entered intValue1 is " + intValue1+ " and strValue1 is " + strValue1);
           
        }catch(NumberFormatException | IOException e) {
            e.printStackTrace();
        }
    }
}