c言語強化訓練作業整理1

1、1つの数(0 x 11 FF)を印刷し、それをポインタとして指向するメモリセルの値を印刷し、指向するメモリセルの値をアドレスとして指向するメモリセルの値を印刷し、このプロセスを繰り返す.

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、       ，         ，              ，                    

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、      0-3000      375           

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

  ，                ，                (200,400)      ，              

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;
		}
	}
}