数据结构之十进制转化8进制及转化2进制

文章后期补充，代码如下(开发工具DEVC++):

#include <iostream>
using namespace std;
#define OK 1
#define ERROR 0
#define OVERFLOW -2
typedef int Status;
typedef struct SNode{
    int data;
    struct SNode *next;
}SNode,*LinkStack;

Status InitStack(LinkStack &S)
{
    S =  NULL;
    return OK;
}

bool StackEmpty(LinkStack S)
{
    if(!S)
        return true;

    return false;
}

Status Push(LinkStack &S, int e)
{
    LinkStack p;
    p = new SNode;
    if(!p)
    {
        return OVERFLOW;
    }
    p->data = e;
    p->next = S;
    S = p;
    return OK;   
}



Status Pop(LinkStack &S, int &e)
{
    LinkStack p;
    if(!S)
        return ERROR;
    e = S->data;
    p = S;
    S = S->next;
    delete p;
    return OK;
}

void conversion(int N)
{
    int e;
    LinkStack S;
    InitStack(S);
    while(N)
    {
        Push(S, N % 8);
        N = N / 8;
    }

    while(!StackEmpty(S))
    {
        Pop(S, e);
        cout << e;
        
    }
}


int main()
{
    int n,e;
    cout << "请输入十进制数字：" << endl;
    cin >> n;
    cout << "十进制" << n << "转化后的8进制为：";
    
    conversion(n);
    
    cout << endl; 
    return 0;
}