新手第一篇-逆波兰计算器

昨天刚刚过了二十岁生日，决定今天开始开始写博客，记录自己的成长，同时也为了以后复习用。回归正题，先上代码

#include<iostream>#include<cctype>using namespace std;typedef double Stack_entry;const int maxstack = 10;enum Error_code{ sucess, overflow, underflow };typedef double Stack_entry;class Stack{public:    Stack();    bool empty()const;//如果Stack为空，返回true，否则返回false    Error_code pop();//如果Stack非空，删除栈顶元素，如Stack为空，则返回underflow（下溢）    Error_code push(const Stack_entry &item);//如果Stack不满，则将元素加到Stack的栈顶，如果Stack是满的，则返回overflow（上溢）    Error_code top(Stack_entry &item)const;//非空的Stack的栈顶元素被复制到item。如果Stack空，返回underflow（下溢）private:    int count;//数组下标    Stack_entry entry[maxstack];//顺序表};Stack::Stack(){    count = 0;}Error_code Stack::push(const Stack_entry &item){    Error_code outcome = sucess;    if (count >= maxstack)        outcome = overflow;    else        entry[count++] = item;    return outcome;}Error_code Stack::pop(){    Error_code outcome = sucess;    if (count == 0)        outcome = underflow;    else        --count;    return outcome;}Error_code Stack::top(Stack_entry &item)const{    Error_code outcome = sucess;    if (count == 0)        outcome = underflow;    else        item = entry[count - 1];    return outcome;}bool Stack::empty()const{    bool outcome = true;    if (count == 0)        outcome = true;    else        outcome = false;    return outcome;}char get_command(){    char command;    bool waiting = true;//循环控制条件    cout << "Select command and press[enter]:";    while (waiting)    {        cin >> command;        command = tolower(command);        if (command == '?' || command == '=' || command == '+' || command == '-' || command == '*' || command == '/' || command == 'q')            waiting = false;        else        {            cout << "Please enter a valid command:" << endl                << "[?]push to stack [=]print top" << endl                << "[+] [-] [*] [/] are arithmetioc operations" << endl                << "[Q]uit" << endl;        }    }    return command;}bool do_command(char command, Stack &numbers){    double p, q;    switch (command)    {    case '?'://将下一位操作数入栈        cout << "Enter a real number:" << flush;        cin >> p;        if (numbers.push(p) == overflow)            cout << "Warning:Sttack full,lost number" << endl;        break;    case '='://输出操作结果        if (numbers.top(p) == underflow)        {            cout << "Stack empty" << endl;        }        else            cout << p << endl;        break;    case'+':        if (numbers.top(p) == underflow)            cout << "Stack empty" << endl;        else        {            numbers.pop();            if (numbers.top(q) == underflow){                cout << "Stack has just one entry" << endl;                numbers.push(p);            }            else            {                numbers.pop();                if (numbers.push(p + q) == overflow)                    cout << "Warning:stack full,lost result" << endl;            }        }        break;    case'-':        if (numbers.top(p) == underflow)            cout << "Stack empty" << endl;        else        {            numbers.pop();            if (numbers.top(q) == underflow)                cout << "Stack has just one entry" << endl;            else            {                numbers.pop();                if (numbers.push(p - q) == overflow)                    cout << "Warning:stack full,lost result" << endl;            }        }        break;    case'*':        if (numbers.top(p) == underflow)            cout << "Stack empty" << endl;        else        {            numbers.pop();            if (numbers.top(q) == underflow)                cout << "Stack has just one entry" << endl;            else            {                numbers.pop();                if (numbers.push(p * q) == overflow)                    cout << "Warning:stack full,lost result" << endl;            }        }        break;    case'/':        if (numbers.top(p) == underflow)            cout << "Stack empty" << endl;        else        {            numbers.pop();            if (numbers.top(q) == underflow)                cout << "Stack has just one entry" << endl;            else            {                numbers.pop();                if (numbers.push(p / q) == overflow)                    cout << "Warning:stack full,lost result" << endl;            }        }        break;    case'q':        cout << "Calculation finished.\n" << endl;        return false;    }    return true;}void introduce(){    cout << "Please enter a valid command:" << endl        << "[?]push to stack [=]print top" << endl        << "[+] [-] [*] [/] are arithmetioc operations" << endl        << "[Q]uit" << endl;}int main(){    Stack stored_numbers;//定义栈类    introduce();//操作说明    while (do_command(get_command(), stored_numbers));}

源自《Data structures and Program Design In c++》

vs2013环境测试结果：

注意：逆波兰计算器是用栈实现的，可以很好的说明栈的使用。在减法测试中，5先入栈，3后入栈，但结果是-2，说明3先出栈，5后出栈，体现了栈的先入后出的特点。

优点：任何表达式，不管多复杂，都可以不用圆括号进行描述

第一次写，难免有不足之处，有任何问题请网友们批评指正