C语言的数组地址数组的遍历与练习

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*
}

 //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.foriSystem.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]);}}