/*
运算符之一:算术运算符
+ - + - * / % (前)++ (后)++ (前)-- (后)-- +
*/
class AriTest {
public static void main(String[] args) {
//除号:/
int num1 = 12;
int num2 = 5;
int result1 = num1 / num2;
System.out.println(result1);//2
原因解释:
这里为什么结果为2呢?
因为 我们的12/5 答案是2.4 但是 这是int / int类型的
系统默认会把它当作int来保存 所以就会出现截断的操作
也就变成了 2
如果想要恢复2.4 可以使用double来保存
double resulte1 = (double)num1 / num2
int result2 = num1 / num2 * num2;
System.out.println(result2);//10
double result3 = num1 / num2;
System.out.println(result3);//2.0
double result4 = num1 / num2 + 0.0;//2.0
double result5 = num1 / (num2 + 0.0);//2.4
double result6 = (double)num1 / num2;//2.4
double result7 = (double)(num1 / num2);//2.0
System.out.println(result5);
System.out.println(result6);
// %:取余运算
//结果的符号与被模数的符号相同
//开发中,经常使用%来判断能否被除尽的情况。
int m1 = 12;
int n1 = 5;
System.out.println("m1 % n1 = " + m1 % n1);
int m2 = -12;
int n2 = 5;
System.out.println("m2 % n2 = " + m2 % n2);
int m3 = 12;
int n3 = -5;
System.out.println("m3 % n3 = " + m3 % n3);
int m4 = -12;
int n4 = -5;
System.out.println("m4 % n4 = " + m4 % n4);
//(前)++ :先自增1,后运算
//(后)++ :先运算,后自增1
int a1 = 10;
int b1 = ++a1;
System.out.println("a1 = " + a1 + ",b1 = " + b1);
int a2 = 10;
int b2 = a2++;
System.out.println("a2 = " + a2 + ",b2 = " + b2);
int a3 = 10;
++a3;//a3++;
int b3 = a3;
//注意点:
short s1 = 10;
//s1 = s1 + 1;//编译失败
//s1 = (short)(s1 + 1);//正确的
s1++;//自增1不会改变本身变量的数据类型
System.out.println(s1);
//问题:
byte bb1 =127;
bb1++;
System.out.println("bb1 = " + bb1); //-128
//(前)-- :先自减1,后运算
//(后)-- :先运算,后自减1
int a4 = 10;
int b4 = a4--;//int b4 = --a4;
System.out.println("a4 = " + a4 + ",b4 = " + b4);
}
}