Java Base language operators

/*use of the + +,--operator: Used alone: placed before and after the operand.            (This usage is common to us) participate in the operation using: Put in front of the operand, the first increment or decrement, and then participate in the operation.                After the operand, it participates in the operation, then increases or is self-reduced. Function: The variable is self-increment 1 or self-minus 1. */ Public classTEXT3 { Public Static voidMain (string[] args) {//define two variables            intx = 3; inty = 4; //Use alone//x + +; //y--;++x; --y; System.out.println ("x:" +x+ ", Y:" +y);//x + + or ++x, when used alone, the output value is the same, here is 4;y--or--y, when used alone, the output value is the same, here is 3System.out.println ("-------------------"); //participate in operations using            intA = 3; intb = 4; //int c = a++; //int d = b--;                        intE = + +A; intf =--b; System.out.println ("A:" +a);//4System.out.println ("B:" +b);//3//System.out.println ("C:" +c);//3//System.out.println ("D:" +d);//4System.out.println ("E:" +e);//4System.out.println ("F:" +f);//3        }}

/*+ +,--Exercises the first question: int a = 10;    int B = 10;    int c = 10;    A = b++;    c =--a;    b = ++a;    A = c--;    Please calculate the value of A,b,c for the second question: int x = 4;    int y = (x + +) + (++x) + (X*10); Please calculate the value of x, Y, respectively*/ Public classTEXT3 { Public Static voidMain (string[] args) {intA = 10; intb = 10; intc = 10; A= b++;//a=10,b=11,c=10c =--a;//a=9,b=11,c=9b = ++a;//a=10,b=10,c=9A = c--;//a=9,b=10,c=8System.out.println ("A:" +a);//9System.out.println ("B:" +b);//TenSystem.out.println ("C:" +c);//8System.out.println ("--------------"); intx = 4; inty = (x + +) + (++x) + (x*10); //4+6+60//x=5,6System.out.println ("x:" +x);//6System.out.println ("y:" +y);// -    }}

/*face Test (possibility): short s=1;s = s+1;        Short s=1;s+=1;                The above two code has no problem, if there is, there is a problem.            Why is there a problem with the second wood?                        The extended assignment operator actually implies a forced type conversion.            s + = 1;            is not equivalent to S = s + 1; Instead, it is equivalent to S = (the data type of s) (S + 1);*/ Public classTEXT4 { Public Static voidMain (string[] args) {//Short s = 1; //s = s + 1; //System.out.println (s);//loss accuracy,                 Shorts = 1; S+ = 1;//seems to be s = s + 1;System.out.println (s);//2    }}

/*bitwise operators: &,|,^,~ <<,>>,>>> Note: To do bit operations, the first thing to do is to convert the data into binary. */ Public classTEXT4 { Public Static voidMain (string[] args) {//&,|,^,~                intA = 3; intb = 4; System.out.println (3 & 4);//0System.out.println (3 | 4);//7System.out.println (3 ^ 4);//7SYSTEM.OUT.PRINTLN (~);//-4    }}/*    Analysis: Because it is a bitwise operation, we must first convert the data into binary. 3 binary: 11 00000000 00000000 00000000 00000011 4 binary: 00000000 00000000 00000000 00000100 & Bit with        Operation: There are 0 0. 00000000 00000000 00000000 00000011 &00000000 00000000 00000000 00000100--------------------------------        ---00000000 00000000 00000000 00000000 The result is: 0 | bit or operation: 1 is 1.        00000000 00000000 00000000 00000011 |00000000 00000000 00000000 00000100-----------------------------------        00000000 00000000 00000000 00000111 The result is: 7 ^ bitwise XOR: The same is 0, the difference is 1. 00000000 00000000 00000000 00000011 &00000000 00000000 00000000 00000100-------------------------------- ---00000000 00000000 00000000 00000111 The result is: 7 ~ bitwise inverse Operator: 0 change to 0 00000000 00000000 00000000 00000011 ~11111111 11111111 11111111 11111100 (complement) complement: 11111111 11111111 11111111 11111100 Anti-code: 1111 1111 11111111 11111111 11111011       Original code: 10000000 00000000 00000000 00000100 The result is:-4 */

/*interview questions: Please realize the exchange of two integer variables Note: In the course of lectures later, I did not explicitly specify the type of data, the default int type. */ Public classTEXT4 { Public Static voidMain (string[] args) {intA = 10; intb = 20; System.out.println ("A:" +a+ ", B:" +b); //Mode 1: Use third-party variables (in development)        /*int c = A;        A = b;        b = C;        System.out.println ("A:" +a+ ", B:" +b);        System.out.println ("------------"); */                //Method 2: Use a different or realization (interview)//Left: A,b,a//Right: a ^ b        /*A = a ^ b; b = a ^ b; A ^ b ^ b = a A = a ^ b;        a^b^a=bsystem.out.println("a:"+a+",b:"+b); */                //Method 3: Adding variables to the procedure        /*A = a + b;//a=30 B = A-c;//b=10 a = a-B;//a=20 System.out.println ("A:" +a+ ", B:" +b); */                //Mode 4: One sentence to take care ofb = (a+b)-(A=B);//b=30-20=10,a=20System.out.println ("A:" +a+ ", B:" +b); }}

Java Base language operators