Hadoop源码分析之RPC机制

#include<stdio.h>
#include<stdlib.h>
int main(int argc, int*argv[])
{
int a[3][2] = { (0, 1), (2, 3), (4, 5) };//看成是一个一维数组，每个元素是大小为2的整型
int*p;
p = a[0];
printf("%d\n", p[0]);
system("pause");
return 0;
}

#include<stdio.h>
#include<stdlib.h>
int main()
{
int a[5][5];
int(*p)[4];
p = a;
printf("a_ptr=%#p\np_ptr=%#p\n",&a[4][2],&p[4][2]);
printf("%p,%d\n", &p[4][2] - &a[4][2], &p[4][2] - &a[4][2]);
system("pause");
return 0;
}

#include<stdio.h>
#include<stdlib.h>
int main()
{
int aa[2][5] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int*ptr1 = (int*)(&aa + 1);
int*ptr2 = (int*)(*(aa + 1));
printf("%d %d\n", *(ptr1 - 1), *(ptr2 - 1));
system("pause");
return 0;
}