operator

1. Overview

Logical operator: the result is a boolean type

&-- yes and (single and)
    1 & 2 want to get the final result to be true, the requirement for 1 and 2 is: both must be true

&&-- yes and (double and / short circuit and)
    1 & & 2, efficient, when the result of 1 is false, 2 will be short circuited

|-- yes or
    1 | 2 if you want the final result to be true, the requirements for 1 and 2 are: 1 or 2 can be true


||-- yes or (double or / short or)
    1 | 2, efficient, when the result of 1 is true, 2 will be short circuited

2. Exercise 1: leap year

Enter the year number to determine if it is a leap year.
Two conditions:
1. Divisible by 4, and not divisible by 100
2. Divisible by 400

package cn.qile.basic;

import java.util.Scanner;

public class Test4_RunNian {
    public static void main(String[] args) {

        //1. Receive the year number entered by the user
        System.out.print("Please enter the year:");
        int year = new Scanner(System.in).nextInt();

        //2. Judge whether year is a normal year or a leap year
        String desc = "Weekday";//Set the default value as the year of the year

        //If (judging condition) {code satisfying the condition}
//     1. Divisible by 4, and not divisible by 100
        if (year % 4 == 0) { //Divisible by 4
            if (year % 100 != 0) { //Cannot be divisible by 100
                desc = "leap year"; //Change desc to leap year
            }
        }
//     2. Divisible by 400
        if (year % 400 == 0) {
            desc = "leap year"; //Change desc to leap year
        }

//     System.out.println("2000 is a leap year");
        //+Concatenate string
        System.out.println(year + "Year is" + desc);

    }
}

Simplified code

Through the relationship of | - yes or (double or / short or)
-- 1 | 2, efficient, when the result of 1 is true, 2 will be short circuited

 //If ((small judgment condition 1 & & small judgment condition 2) | big judgment condition 2) {change desc value to leap year}
       if(( year%4==0  &&  year%100 != 0 )  ||  year%400 == 0  ){
           desc="leap year";
       }

3. Exercise 2: increase and decrease

package cn.qile.basic;

//Test auto increment + + auto decrement--
public class Test5_ZiZeng {
    public static void main(String[] args) {
        //Symbol before, change before use
        //Symbols are used first and then changed
        int a = 1;
        System.out.println(a++);//1

        int b = 1;
        System.out.println(++b);//2
        System.out.println(++b+a+b++);//8，3+2+3

        //TODO self reduction
        int m = 1;
        System.out.println(m--);//1

        int n = 1;
        System.out.println(--n);//0
        System.out.println(--m-n-m--);//0，-1-0-(-1)
    }
}

4. Ternary operation

package cn.qile.basic;

import java.util.Scanner;

//Ternary operator 
public class Test6_Max {
    public static void main(String[] args) {

        //The big value of two numbers
        int a = 10;
        int b = 6;

        //Three yuan 1? 2: 3
        //The final result is 2 or 3. It depends on the result of 1. If 1 is established, 2 will be obtained. If 1 is not established, 3 will be obtained
        int max = a > b ? a : b ;
        System.out.println(max);

        //Receive the three numbers entered by the keyboard, and take out the maximum value
        System.out.print("Please enter m =");
        int m = new Scanner(System.in).nextInt();
        System.out.print("Please enter n =");
        int n = new Scanner(System.in).nextInt();
        System.out.print("Please enter o =");
        int o = new Scanner(System.in).nextInt();

        //1. Define the maximum value of variable record, take the ratio of m and n, and submit the result to the variable for saving
        int result = m > n ? m : n;

        //2. Take the ratio of result and o, and the maximum value will be given to maxValue for saving
//     result = result > o ? result : o ;
        int maxValue = result > o ? result : o ;
        System.out.println(maxValue);

        //TODO optimization
        //Take the ratio of m to n? The ratio of m to o: the ratio of n to o
        int result2 = m > n ? (m > o ? m : o ) : (n > o ? n : o );
        System.out.println(result2);
    }

}