int a[6]={1,2,3,4,5,6},*p; p=a+2; k=*(p+2);h=*(p-2)+p[1]

int a[5]={2,4,6,8,10},*P,* *k; p=a; k=&p; printf("%d",*(p++) { int a[]={1,2,3,4,5,6}; int*p; p=a; printf("%d\n",*p); prin #include "stdio.h" main() { int a []={1,2,3,4,5},*p;p=a;*(p+ main() {int a[5]={2,4,6,8,10},*p,**k; p=a; k=&p; printf("%d" main() {int a【5】＝｛2,4,6,8,10｝,*p,**k; p=a;k=&p; C语言 k=&p main(){ int a[5]={2,4,5,6,10},*p,**k;p=a; k=&p; pri int a[]={1,2,3,4,5,6,7,8,9},*p; for(p=a,p int a[3][3]={{1,2,3},{4,5,6},{7,8,9}}; int **p; p=(int**)a; int a[2][3]={0,1,2,3},*p; p=&a[2]; *--p int a[3][5]={{1,2},{6,4},{3,4,5}}； int (*p)[5] = a; 求*(*p+1) #include"stdio.h" fun(int k,int *p) {int a,b;if(k==1||k==2)* int[][4]={1,2,3,4,5,6,7,8};int(*p)[4]=a;则表达式**（p+1）-*(*p+1)的