为什么 getline 没有在我的程序顶部工作，而是在另一个函数中工作？

Question

我正在开发一个循环程序，直到用户输入

y

或

Y

以外的字符以停止它。我正在使用

std::getline()

，因为我需要在另一个函数中从用户那里获取整行文本，并且它在我的程序的那个函数中运行良好。但是，当要求用户继续或停止角色时，它不起作用。我遇到编译器错误：

调用'getline'没有匹配函数

我正在使用

std::getline()

，因为

std::cin

和

std::getline()

在一起会导致不需要的行为。我不知道如何解决这个问题，希望能得到一些帮助。

#include <iostream>
#include <string>
#include <stack>
using namespace std;

static int ASCII_OFFSET = 48;
//48 is the ASCII value for char '0', so we take the ASCII value of char '0' and subtract 48 to get the integer 0

void postCalculator();

int main()
{
    char again = 'y';
    while(again == 'y'||again == 'Y')
    {
        postCalculator();
        cout << "Type 'Y' or 'y' to continue, or type any other letter to quit:";
        getline(cin, again);
    }
}

/////
///// the line:       getline(cin, again);         is the issue
/////

void postCalculator()
{
    string equation;
    stack<int> stack1;
    int a, b, c, holder;

    cout << "Please enter the RPN expression to be calculated:" << endl;
    getline(cin, equation);

    for(int i = 0; i < equation.length(); i++)
    {
        if(equation[i] >= '0' && equation[i] <= '9')
        {
            holder = static_cast<int>(equation[i]) - ASCII_OFFSET;
            cout << "Token = " << holder << ", ";
            stack1.push(holder);
            cout << "Push " << holder << endl;
        }
        else if(i == ' ')
        {
            continue;
        }
        else if(equation[i] == ':')
        {
            break;
        }

        if(equation[i] == '-')
        {
            cout << "Token = -, ";
            a = stack1.top();
            stack1.pop();
            cout << "Pop " << a << " ";
            b = stack1.top();
            stack1.pop();
            cout << "Pop " << b << " ";

            c = b - a;

            stack1.push(c);
            cout << "Push " << c << endl;
        }
        else if(equation[i] == '*')
        {
            cout << "Token = *, ";
            a = stack1.top();
            stack1.pop();
            cout << "Pop " << a << " ";
            b = stack1.top();
            stack1.pop();
            cout << "Pop " << b << " ";

            c = b * a;

            stack1.push(c);
            cout << "Push " << c << endl;
        }
        else if(equation[i] == '/')
        {
            cout << "Token = /, ";
            a = stack1.top();
            stack1.pop();
            cout << "Pop " << a << " ";
            b = stack1.top();
            stack1.pop();
            cout << "Pop " << b << " ";

            c = b / a;

            stack1.push(c);
            cout << "Push " << c << endl;
        }
        else if(equation[i] == '+')
        {
            cout << "Token = +, ";
            a = stack1.top();
            stack1.pop();
            cout << "Pop " << a << " ";
            b = stack1.top();
            stack1.pop();
            cout << "Pop " << b << " ";

            c = b + a;

            stack1.push(c);
            cout << "Push " << c << endl;
        }
    }
    cout << "Token = Pop " << stack1.top() << endl << endl;
    stack1.pop();
}

我尝试使用

std::cin

代替

std::getline()

，以及

std::cin.clear()

，虽然我不完全理解这种行为，但没有成功。

Answer 1

您不能使用

char

读取单个

std::getline()

，它仅适用于

std::(basic_)string

类型。所以，你将不得不：

• 把

  char again

  改成

  std::string again

  ，例如：

int main()
{
    ...
    string again;
    do
    {
        ...
    }
    while (getline(cin, again) &&
          (again == "y" || again == "Y"));
    ...
}

• 用

  operator>>

  代替

  std::getline()

  ，例如：

...
#include <limits>
...

int main()
{
    ...
    char again;
    do
    {
        ...
        if (!(cin >> again)) break;
        cin.ignore(numeric_limits<streamsize>::max(), '\n');
    }
    while (again == 'y' || again == 'Y');
    ...
}