标签：strong bsp 算术 oid system ble 编译 符号 改变
/*
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
int result2=num1/num2*num2;
System.out.println(result2);
double result3=num1/num2;
System.out.println(result3);
double result4=num1/num2+0.0;
double result5=num1/(num2+0.0);
double result6=(double)num1/num2;
double result7=(double)(num1/num2);
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chu‘jin
int m1=12;
int n1=5;
System.out.println("m1 % n1 ="+m1%n1);
int m2=-12;
int n2=5;
System.out.println("m1 % n1 ="+m2%n2);
int m3=12;
int n3=-5;
System.out.println("m1 % n1 ="+m3%n3);
int m4=-12;
int n4=-5;
System.out.println("m1 % n1 ="+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不会改变本身变量大的数据类型
//问题:
byte b1=127;
bb1++;
System.out.println("bb1="+bb1);
//(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);
}
}
自加 自减
标签：strong bsp 算术 oid system ble 编译 符号 改变
原文地址：https://www.cnblogs.com/xuezha/p/12088463.html
