指针数组经典例题 总结解析

```c
int main()
{
	int a[4] = { 1, 2, 3, 4 };
	int *ptr1 = (int *)(&a + 1);
	int *ptr2 = (int *)((int)a + 1);// 若a=ox100   (int)a+1=101
	printf( "%x,%x", ptr1[-1], *ptr2);// 4   2000000
	return 0;
}

int main()
{
	int a[5][5];
	int(*p)[4]  = (int (*)[4])a;
	printf( "%p,%d\n", &p[4][2] - &a[4][2],        &p[4][2] - &a[4][2]);//
	return 0;
}

int main()
{
	int aa[2][5] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
	int *ptr1 = (int *)(&aa + 1);//&aa  aa代表整个数组  取出整个数组的地址
	int *ptr2 = (int *)(*(aa + 1));//aa+1  第二行地址
	printf( "%d,%d", *(ptr1 - 1), *(ptr2 - 1));// 5   10
	return 0;
}

int main()
{
	char *c[] = {"ENTER","NEW","POINT","FIRST"};
	char**cp[] = {c+3,c+2,c+1,c};
	char***cpp = cp;
	printf("%s\n", **++cpp); //POINT
	printf("%s\n", *--*++cpp+3);// ER   *(--*(++CPP))+3     
	printf("%s\n", *cpp[-2]+3);// *(*(CPP-2))+3   ST
	printf("%s\n", cpp[-1][-1]+1);//  *(*(CPP-1)-1)+1  //EW
	return 0;
}

char *myString()
{
	char buffer[6] = {0};
	char *s = "Hello World!";
	int i = 0;
	for (i = 0; i < sizeof(buffer) - 1; i++)
	{
		buffer[i] = *(s + i);
	}
	return buffer;//return  char类型
}
int main()
{
	printf("%s\n", myString());//乱码
	return 0;
}

//局部变量分配的内存在使用完之后应该及时释放