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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 "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 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 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。 
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const uchar table[256] = {... ...};
result = *(table+convchar);