栈练习——进制转换

#include<iostream>
using namespace std;

#define OVERFLOW -2
#define OK 1
#define ERROR -1

typedef int Status;
typedef int SElemType;

typedef struct StackNode {
	SElemType data;
	struct StackNode* next;
}StackNode, * LinkStack;

// 链栈初始化
void InitStack(LinkStack& S) {
	S = NULL;
}

// 判断链栈是否为空
Status StackEmpty(LinkStack S) {
	if (S == NULL) return 1;
	else return 0;
}

// 入栈
Status Push(LinkStack& S, SElemType e) {
	LinkStack p;
	p = new StackNode;
	if (!p) exit(OVERFLOW);
	p->data = e;
	p->next = S;
	S = p;
	return OK;
}

// 出栈
Status Pop(LinkStack& S, SElemType& e) {
	if (StackEmpty(S)) return ERROR;
	LinkStack p;
	e = S->data;
	p = S;
	S = S->next;
	delete p;
	p = NULL;
	return OK;
}

/*---------------进制转换---------------*/
void Invert(LinkStack &S, int n) {
	SElemType e;
	int m;
	cout << "请输入要转换的进制: ";
	cin >> m;
	while (n) {
		Push(S, n % m);
		n = n / m;
	}

	cout << "转换后的数字为: ";
	while (!StackEmpty(S)) {
		Pop(S, e);
		cout << e;
	}
	cout << endl;
}

int main()
{
	int n;
	LinkStack S;
	InitStack(S);
	cout << "请输入要转换的数字: ";
	cin >> n;
	Invert(S, n);
	return 0;
}

  

 

  
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请输入要转换的数字: 1024
请输入要转换的进制: 2
转换后的数字为: 10000000000

  

 

  
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文章来源: ruochen.blog.csdn.net，作者：若尘，版权归原作者所有，如需转载，请联系作者。

原文链接：ruochen.blog.csdn.net/article/details/102800243