简单静态栈实现

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简单栈的实现

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//Stack.cppenum Error_code {SUCCESS, UNDERFLOW, OVERFLOW};typedef int Stack_entry;const int maxstack = 5; //使用小容量方便测试class Stack {public:    Stack();    bool empty() const;  //状态不可改    Error_code pop();  //为什么弹出的状态没有const???    Error_code top(Stack_entry& item) const; //就是说这个top不能调用pop(),只能调用const函数，例如empty()    Error_code push(const Stack_entry &item);private:    int count;    Stack_entry entry[maxstack];};Stack::Stack() {    count = 0;}bool Stack::empty() const {    if (count == 0) {        return true;    } else {        return false;    }}Error_code Stack::pop() {    if (count == 0) {        return UNDERFLOW;    } else {        --count;        return SUCCESS;    }}Error_code Stack::push(const Stack_entry& item) {    if (count >= maxstack) {        return OVERFLOW;    } else {        entry[count] = item;        ++count;        return SUCCESS;    }}Error_code Stack::top(Stack_entry &item) const {    if (count == 0) {        return UNDERFLOW;    } else {        item = entry[count-1]; //top()只返回当前顶端的值，count的值应该不变        return SUCCESS;    }}

//main.cpp#include <iostream>#include "Stack.h"using namespace std;int main() {    Stack myStack;    int temp;    myStack.push(1);    myStack.push(3);    myStack.push(4);    myStack.push(8);    myStack.push(2);  //2,8,4,3,1    myStack.push(0);  //2,8,4,3,1    myStack.top(temp);    cout << temp << endl;    myStack.pop();  //8,4,3,1    myStack.top(temp);    cout << temp << endl;    myStack.push(9);  //9,8,4,3,1    myStack.top(temp);    cout << temp << endl;    myStack.pop();  //8,4,3,1    myStack.top(temp);    cout << temp << endl;    myStack.pop();  //4,3,1    myStack.pop();  //3,1    myStack.pop();  //1    myStack.pop();  //(empty)    myStack.pop();  //(empty)    myStack.top(temp);  //输出为8，说明了其实top()函数最好还是要有Error_code的判断    cout << temp << endl;    return 0;}

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