c++ - 如何对堆栈中的所有项目求和？后缀计算器

Question

我正在制作一个后缀计算器，我想为其添加一个新功能。如何将堆栈中的所有项目相加，然后从堆栈中删除用于求和的项目，然后将总和推回堆栈？所有这一切都应该通过按下命令“s”来实现。

我试过这样做，但它总是只打印 0。

    case 's':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                while (!numbers.empty()) {
                    sum += numbers.top().second;

                }
                numbers.push(sum);
                cout << sum << endl;

            }
        }
        break;

这是我的 main.cpp 的一些代码。

#include <iostream>
#include "Stack.h"
#include <cmath>
using namespace std;

void help()
{
    cout
        << "[?]push to stack [=]print top [x]exchange [s]sum [a]average" << endl
        << "[+] [-] [*] [/] [%] [^] [v] are arithmetic operations" << endl
        << "[Q]uit." << endl;
}

char get_command()
{
    char command;
    bool waiting = true;
    cout 
        << "Select command and press <Enter>:" << endl;

    while (waiting) {
        cin >> command;
        command = tolower(command);
        if (command == '?' || command == '=' || command == '+' ||
            command == '-' || command == '*' || command == '/' ||
            command == 'q' || command == 'x' || command == 's' || 
            command == 'a' || command == '%' || command == '^' || 
            command == 'v') waiting = false;


        else {
            cout << "Please enter a valid command:" << endl;
        }
    }
    return command;
}


bool do_command(char command, Stack& numbers)

{
    double p, q, g, r;
    double sum = 0;
    switch (command) {
    case '?':
        cout << "Enter a real number: " << flush;
        cin >> p;
        if (numbers.push(p) == overflow)
            cout << "Warning: Stack full, lost number" << endl;
        break;

    case '=':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else
            cout << p << endl;
        break;

    case '+':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                numbers.pop();
                if (numbers.push(q + p) == overflow)
                    cout << "Warning: Stack full, lost result" << endl;
            }
        }
        break;

    case '-':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                numbers.pop();
                if (numbers.push(q - p) == overflow)
                    cout << "Warning: Stack full, lost result" << endl;
            }
        }
        break;

    case '*':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                numbers.pop();
                if (numbers.push(q * p) == overflow)
                    cout << "Warning: Stack full, lost result" << endl;
            }
        }
        break;

    case '/':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                numbers.pop();
                if (numbers.push(q / p) == overflow)
                    cout << "Warning: Stack full, lost result" << endl;
            }
        }
        break;

    case 'x':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                numbers.pop();
                numbers.pop();
                numbers.push(p);
                numbers.push(q);
            }
        }
        break;

    case 's':
        if (numbers.top().first == underflow)
            cout << "Stack empty" << endl;
        else {
            numbers.pop();
            if (numbers.top().first == underflow) {
                cout << "Stack has just one entry" << endl;
                numbers.push(p);
            }

            else {
                while (!numbers.empty()) {
                    sum += numbers.top().second;

                }
                numbers.push(sum);
                cout << sum << endl;

            }
        }
        break;

这是我的 Stack.cpp。

Error_code Stack::push(const Stack_entry& item)
/*
Pre:  None.
Post: If the Stack is not full, item is added to the top
      of the Stack.  If the Stack is full,
      an Error_code of overflow is returned and the Stack is left unchanged.
*/

{
    Error_code outcome = success;
    if (count >= maxstack)
        outcome = overflow;
    else
        entry[count++] = item;
    return outcome;
}


Error_code Stack::pop()
/*
Pre:  None.
Post: If the Stack is not empty, the top of
      the Stack is removed.  If the Stack
      is empty, an Error_code of underflow is returned.
*/

{
    Error_code outcome = success;
    if (count == 0)
        outcome = underflow;
    else --count;
    return outcome;
}


pair<Error_code, int> Stack::top() const
{
    pair<Error_code, int> p;
    Error_code outcome = success;
    if (count == 0) 
        p.first = underflow;
        p.second = 0;
        return p;

    p.first = outcome;
    p.second = entry[count - 1];
    return p;
}

这是我的 Stack.h

#include "Utility.h"

typedef double Stack_entry;

const int maxstack = 10;

class Stack {
public:
    Stack();
    bool empty() const;
    Error_code pop();
    //Error_code top(Stack_entry& item) const;
    Error_code push(const Stack_entry& item);
    //Error_code pop_top(Stack& s, Stack_entry& item);
    double size() const; 
    std::pair<Error_code, int> top() const;

private:
    int count;
    Stack_entry entry[maxstack];
};

我希望它将堆栈中的所有数字加在一起，然后删除它们，然后将其推回堆栈。我真的很感激任何帮助。提前致谢。

Answer 1

pop 不像 C++ 中的其他语言那样返回值。您需要使用 top 函数来返回堆栈中的顶部元素。您可以将该元素添加到 sum，然后使用 pop 将其删除。您可以按如下方式实现它:

while (!numbers.empty()) 
{
    sum += numbers.top();
    numbers.pop()
}

对于 Op 的另一个请求:

将 top 的实现更改为:

std::pair<Error_code,int> top() const
    {
        std::pair<Error_code,int> p;
        Error_code outcome = success;
        if (count == 0)
            p.first = underflow;
            p.second = 0;
            return p;

        p.first = outcome;
        p.second = entry[count -1];
        return p;
    }

你的第一个问题的解决方案变成了:

while (!numbers.empty()) 
    {
        sum += numbers.top().second;
        numbers.pop()
    }

如果你想要结果，你可以做numbers.top().first。就这样。