1.int main(void)
{
    int a[5] = { 10,20,30,40,50 };//数组间的元素地址相连的
    int* p;
    printf("%d\n", &a[0]);
    printf("%d\n", &a[1]);
    printf("%d\n", &a[2]);
    printf("%d\n", &a[3]);
    printf("%d\n", &a[4]); 
    printf("a代表的地址：%d\n", a);//数组首个元素的地址
    printf("a+1代表的地址：%d\n", a+1);//偏移数组
    printf("a+2代表的地址：%d\n", a+2);
    printf("a+3代表的地址：%d\n", a+3);
    printf("a+4代表的地址：%d\n", a+4);
    printf("a[4]代表的地址：%d\n", &a[2]+2);
    printf("*(a+2)的值：%d\n", *(a+2));//调用数组元素用*
}

2.

 //数组的遍历与练习
   /* public static void main(String[] args) {
        int []arr={1,2,3,4,5};
        for (int i = 0; i <= 4; i++) {
            System.out.println(arr[i]);
        }
        System.out.println(arr.length);//数组的长度
        for (int i = 0; i < arr.length; i++) {//arr.fori
            System.out.println(arr[i]);
        }
    }*/
    //遍历数组并求和
    /*public static void main(String[] args) {
int sum=0;
int []arr={1,2,3,4,5};
        for (int i = 0; i < arr.length; i++) {
            sum=sum+arr[i];
        }
        System.out.println(sum);
    }*/
   /* public static void main(String[] args) {
        int []arr={1,2,3,4,5,6,7,8,9,10};
        int num=0;
        for (int i = 0; i < arr.length; i++) {
            if(arr[i]%3==0){
                System.out.println(arr[i]);
                num++;
            }
        }
        System.out.println(num);
    }*/
    public static void main(String[] args) {
        int []arr={1,2,3,4,5,6,7,8,9,10};
        for (int i = 0; i < arr.length; i++) {
            if(arr[i]%2==0){
                arr[i]=arr[i]/2;
            } else  {
                arr[i]=arr[i]*2;
            }
        }
        for (int i = 0; i < arr.length; i++) {
            System.out.println(arr[i]);
        }
    }