进制转换代码练习(支持长二进制)

进制转换代码(支持长二进制)：

Blist.h

Blist.h
#pragma once

#define ok 1
#define error 0
#define true 1
#define false 0
#define overflow -2
#define MAXSIZE 100

typedef int Status;

typedef struct BNode {
    int data;
    struct BNode *pre, *next;
}BNode,*Position;

typedef struct {
    BNode *head;
    BNode *tail;
}Blist;

建立链表函数：

#include "Blist.h"
#include <stdio.h>

Status InitBlist(Blist &BL) {//建立双链表
    BNode *p, *q;
    p = (BNode *)malloc(sizeof(BNode));
    if (!p) exit(overflow);
    p->data = NULL;
    q = (BNode *)malloc(sizeof(BNode));
    if (!q) exit(overflow);
    q->data = NULL;
    p->next = p->pre = q;
    q->next = q->pre = p;
    BL.head = p; BL.tail = q;
    return ok;
}

Status AddElem(Blist &BL, long e) {//插入e到双链表中
    BNode *p;
    p = (BNode *)malloc(sizeof(BNode));
    if (!p) exit(overflow);
    p->data = e;
    p->next = BL.head->next; BL.head->next = p;//头尾链接
    p->pre = p->next->pre; p->next->pre = p;
}

void PrtBlist(Blist BL, char p) {//打印双链表
    Position q;
    switch (p) {//通过p判断从头打印还是从尾打印
    case 'H': {//从头
        q = BL.head->next;
        while (q->next != BL.head) {//
            printf("%ld", q->data);
            q = q->next;
        }//
        break;
    }//
    case 'T': {//从尾
        q = BL.tail->pre;
        while (q->pre != BL.tail) {//
            printf("%d", q->data);
            q = q->pre;
        }//
    }//
    }//
}//

主区：

#include <stdio.h>
#include <math.h>
#include "Blist.h"

//-------函数声明
long BtoD(long);
long DtoB(long);
long Len(long);

//--------外部函数声明
extern Status InitBlist(Blist &BL);
extern Status AddElem(Blist &BL, long e);
extern void PrtBlist(Blist BL, char p);


//-----------主函数
int main() {
    long B;
    char J;
    long (*p)(long);//函数指针
    scanf("%c%ld", &J,&B);
    switch (J) {
    case'B':p = BtoD; printf("%ld", (*p)(B));  break;
    case'D':p = DtoB;  (*p)(B); break;
    }
    
    return 0;
}

//-----------函数定义
long DtoB(long a) {//十进制转二进制
    Blist BL;
    int i;
    long r=a,p,H=0;
    InitBlist(BL);
    for (i = 0; r != 0; i++) {
        p = r % 2;//先求余
        r = r/2;//再求整除
        //H += p * pow(10, i);//相加
        AddElem(BL, p);
    }
    PrtBlist(BL,'H');
    return ok;
}

long BtoD(long D) {//二进制转十进制
    int i,H=0,r=D;
    int le,p;
    le = Len(D);
    for (i = le; i >= 0; i--) {
        p = r / pow(10, i-1);//求位次上的数，如1234的1
        r = r - p*pow(10, i - 1);//求退一位的余，如从1234求得234
        H += p * pow(2,i-1);
    }
    return H;
}

long Len(long NUM) {
    int i=0;
    do{
        NUM = NUM / 10;
        i++;
    } while (NUM!=0);
    return i;
}
/*心得：
1.可以通过注释标记判断括号匹配情况；
2.加入双链接结构使程序复杂，但是能够输出长二进制
*/