public static void main(String[] args){
    sum(1, 5, 100) //返回106
    sum(1, 6) //返回6
}

public static int sum(int... nums){
    int res = 0;
    for(int i = 0; i < nums.length; i++){
        res += nums[i];
    }
    return res;
}// int... 表示接受的是可变参数，类型是int，即可接受多个int(0-多)
 // nums 可以当作数组

public void f1(double... nums, String str)// 错误
public void f2(String str, double... nums)// 正确

public void f3(int... nums1, double... nums2)// 错误

//令方法：calculate
//下面四个calculate构成重载
public int calculate(int n1, int n2){
    return n1 + n2;
}
public double calculate(int n1, double n2){
    return n1 + n2;
}
public double calculate(double n1, int n2){
    return n1 + n2;
}
public int calculate(int n1, int n2, int n3){
    return n1 + n2 + n3;
}

public int calculate(int n1, int n2, int n3){
    return n1 + n2 + n3;
}
//不构成重载，因为和返回类型无关
public void calculate(int n1, int n2, int n3){
    int n = n1+n2+n3;
}

int a = 10;
int b = 99;
int result = a>b ? a++ :b--;//result = 99，但是a = 10, b = 98

int a = 3;
int b = 8;
int c = a > b ? 1.1 : 3.4;//错误的
double d = a > b ? a : b + 3;//可以的

int max = (n1 > n2 ? n1 : n2) > n3 ? (n1 > n2 ? n1 : n2) : n3;

int n1 = 10;
float d1 = n1 + 1.1;//错误，n1 + 1.1 => 结果是double
double d1 = n1 + 1.1;//对
float d1 = n1 + 1.1f;//对，将 1.1 double转float

int n2 = 1.1;//错误 double -> int

byte b1 = 10;//对，-128 - 127
int n2 = 1;
byte b2 = n2;//错误，如果是变量赋值，判断类型
char c1 = b1;//错误，原因 byte 不能自动转换成 char

byte b2 = 1;
short s1 = 1;
short s2 = b2 + s1;//错，b2 + s1 => int
int s2 = b2 + s1;
byte b4 = b2 + b3;//错， b2 + b3 => int

byte b4 = 1;
short s3 = 100;
int num200 = 1;
double num300 = 1.1;
double num500 = b4 + s3 + num200 + num300;