用指针实现高低位倒序，疯了吧？

昨晚在微信群看到一个读者发的面试题目，从网上截图出来的，我百思不得其解，题目如图。

幸好，我学过栈

然后我写了个小程序

第一个方法比较笨，当我写完自己的代码后，看到有同学发了自己的代码，我赶紧就发了个红包，一个是为了鼓励大家多讨论问题，一个是为了赞扬这样的行为。

第一个方法就是取出每一个bit，压栈，然后再出栈，你想怎么排序都没有问题。

#include "stdio.h"
#include "stdlib.h"

struct List{
	int data;
	struct List * next;
};

struct Stack{
	struct List *head;
	int size;
};


struct Stack * StackInit(void)
{
	struct Stack *stack = NULL;
	stack =  (struct Stack*)malloc(sizeof(struct Stack));
	stack->head = (struct List *)malloc(sizeof(struct List));
	stack->head->next = NULL;
	stack->size = 0;
	return stack;
}

int StackPush(struct Stack *stack,int data)
{
	struct List *tmp = (struct List *)malloc(sizeof(struct List));
	tmp->data = data;
	tmp->next = stack->head->next;
	stack->head->next = tmp;
	stack->size++;
	//printf("push:%d \n",data);
	return 0;
}

int IsStackEmpty(struct Stack *stack)
{
	if(stack->head->next == NULL){
		//printf("stack null\n");
		return 1;
	}else{
		//printf("stack is not null\n");
		return 0;
	}
}

int StackPop(struct Stack *stack,int *data)
{
	struct List *tmp = NULL;
	if(IsStackEmpty(stack))
		return -1;
		
	tmp = stack->head->next;
	*data = tmp->data;
	stack->head->next = tmp->next;
	stack->size--;
	if(tmp!=NULL)
		free(tmp);
	
	//printf("pop:%d \n",*data);
	return 0;
}


int main(void)
{
	int i = 0;
	unsigned char a = 0x71;
	unsigned char b = 0;
	
	struct Stack *stack = NULL;
	stack = StackInit();

	printf("a:0x%x\n",a);
	for(i = 0;i<8;i++)
	{

		if(a&0x10) StackPush(stack,1);
		else       StackPush(stack,0);
		if(a&0x20) StackPush(stack,1);
		else       StackPush(stack,0);
		if(a&0x40) StackPush(stack,1);
		else       StackPush(stack,0);
		if(a&0x80) StackPush(stack,1);
		else       StackPush(stack,0);

		if(a&0x01) StackPush(stack,1);
		else       StackPush(stack,0);		
		if(a&0x02) StackPush(stack,1);
		else       StackPush(stack,0);
		if(a&0x04) StackPush(stack,1);
		else       StackPush(stack,0);	
		if(a&0x08) StackPush(stack,1);
		else       StackPush(stack,0);					
	}
	for(i = 0;i<8;i++)
	{
		int d = 0;
		StackPop(stack,&d);
		if(i == 0)
			if(d == 1) b |= 0x80;
		if(i == 1)
			if(d == 1) b |= 0x40;
		
		if(i == 2)
			if(d == 1) b |= 0x20;
		
		if(i == 3)
			if(d == 1) b |= 0x10;

		if(i == 4)
			if(d == 1) b |= 0x08;

		if(i == 5)
			if(d == 1) b |= 0x04;
	 
		if(i == 6)
			if(d == 1) b |= 0x02;

		if(i == 7)
			if(d == 1) b |= 0x01;

	}
	
	printf("\nb:0x%x\n",b);
	return 0;
}

执行如下：

除了用栈来实现，群里有位同学的答案让我眼前一亮

#include <stdio.h>

struct charbits{
 unsigned char data1 : 4;
 unsigned char data2 : 4;
};


union dataChange{
 struct charbits charBits;
 unsigned char data;
}datchange,*p;

int main(){

 unsigned char temp;

 p = &datchange;

 datchange.data = 0x34;

 temp = p->charBits.data1;

 p->charBits.data1 = p->charBits.data2;

 p->charBits.data2 = temp;

 printf("data is %x", datchange.data);

 return 0;
}

完美演示了位域、共用体、结构体、指针的几个知识点~

测试下位域的作用

#include <stdio.h>

struct charbits{
 unsigned char data1 : 1;
 unsigned char data2 : 3;
};


struct charbits1{
 unsigned char data1;
 unsigned char data2;
};

int main(){

 struct charbits a;
 struct charbits1 b;
 printf("%d %d\n",sizeof(a),sizeof(b));
 
 /*赋值1看看*/
 a.data1 = 1;
 printf("%d\n",a.data1);
 /*赋值3看看*/
 a.data1 = 3;
 printf("%d\n",a.data1);
 
 /*赋值3看看*/
 a.data2 = 3;
 printf("%d\n",a.data2);

 /*赋值7看看*/
 a.data2 = 7;
 printf("%d\n",a.data2);
 
 
 /*赋值8看看*/
 a.data2 = 8;
 printf("%d\n",a.data2);
 return 0;
}

data1只能存 0和1，如果存其他值的话就会溢出变成0。

data2是 3bits ，只能存 0~7，我们存 8 进去，就溢出变成 0。

本文授权转载自公众号“嵌入式Linux”，作者写篮球的代码球痴

-END-

推荐阅读

【01】再谈指针：大佬给你拨开 C指针 的云雾【02】嵌入式编程中的复杂指针的使用【03】C语言指针用得好犹如神助！这些使用技巧值得收藏【04】值得收藏的 C语言 指针讲解文章【05】指针，很难吗？| 工程师给你详细解释！