c语言强化训练作业整理1

1、将一个数（0x11FF）打印，再将它作为指针打印指向的内存单元中的值，再将所指向的内存单元的值再作为地址打印指向的内存单元的值，重复这个过程。

main() 
{
	int p = 0x11ff;
	char ch = 0;
	while (ch != 'q')
	{
		printf("p = %4x  \t*p = %4x\n",p,(int)(*(int*)p));
		p = (int *)(*(int*)p);
		ch = getch();
	}
}

2、编写一个结构体，打印结构体的首地址，再打印结构体中每个数的首地址，然后将结构体按照一个字节一个字节打印出来

main()
{
	int i;
	
	struct st
	{
		char c1;
		char c2;
		int i1;
		int i2;
	};
	
	struct st st;
	st.c1 = 'a';
	st.c2 = 'b';
	st.i1 = 1;
	st.i2 = 2;
	printf("&st = %x\n", &st);
	printf("&c1 = %x\n", &st.c1);
	printf("&c2 = %x\n", &st.c2);
	printf("&i1 = %x\n", &st.i1);
	printf("&i2 = %x\n\n", &st.i2);
	
	for (i=0; i<sizeof(st); i++)
	{
		printf("%x  ",(*(char *)(&st.c1+i)));
	}
}

3、将偏移地址为0-3000的内存段看做375个具有以下结构的结构体

struct st
{
	unsigned char int1;
	unsigned char int2;
	unsigned char int3;
	struct st far * pst;
};

要求，将这段内存复制到一个结构体数组中，从数组中选出结构体的三个值的和在(200,400)范围内的变量，通过结构体中的指针连接成链表

struct st
{
	unsigned char int1;
	unsigned char int2;
	unsigned char int3;
	struct st far * pst;
};

main() 
{
	struct st starray[375];
	int ii,sum;
	
	struct st far * pHead = 0;
	struct st far * pNow = 0;
	
	/*copy*/
	for (ii=0; ii<375; ii++)
	{
		starray[ii] = *((struct st *)(0 + ii * sizeof(struct st)));
	}
	
	/*count*/
	for (ii=0; ii<375; ii++)
	{
		sum = starray[ii].int1 + starray[ii].int2 + starray[ii].int3;
		if (sum < 400 && sum > 200)
		{
			if (pNow == 0)
			{
				pHead = &(starray[ii]);
				pNow = pHead;
			}
			else
			{
				pNow->pst = &(starray[ii]);
				pNow = pNow->pst;
			}
		}
	}
	pNow->pst = 0;
	
	/*output*/
	pNow = pHead;
	sum = 0;
	while (pNow) 
	{
		printf("int1 = %d,\tint2 = %d,\tint3 = %d,\tintSum = %d,\tpNext = %x%x\n",pNow->int1,
				pNow->int2,pNow->int3,pNow->int1+pNow->int2+pNow->int3,pNow->pst);
		pNow = pNow->pst;
		sum ++;
		
		if (sum > 10)
		{
			printf("----------------------------------------------------------------------------\n");
			getch();
			sum = 0;
		}
	}
}