C++ 循环 - 流错误？

Question

在我的编程课上，我们的任务是编写一个程序，将单词或短语转换为电话号码，评估每个字符并将其翻译成相应的数字。

这是目前的代码:

#include <iostream>
#include <string>

using namespace std;

int main()
{
    char letter;
    int noOfLetters;
    char response;

    cout << "Enter Y/y to convert a telephone number " 
         << "form letters to digits.\n"
         << "Enter any other letter to terminate the program: ";

    cin >> response;
    cout << endl;

    while (response == 'Y' || response == 'y')
    {

        cout << "Enter a telephone number using letters: ";
        cin >> letter;

        noOfLetters = 0;

        cout << "The corresponding telephone number is: ";

        while (noOfLetters != 7)
        {
            //cout << "[" << noOfLetters << "]";
            noOfLetters++;

            switch (toupper(letter))
            {
            case 'A':
            case 'B':
            case 'C':
                cout << 2;
                break;
            case 'D':
            case 'E':
            case 'F':
                cout << 3;
                break;
            case 'G':
            case 'H':
            case 'I':
                cout << 4;
                break;
            case 'J':
            case 'K':
            case 'L':
                cout << 5;
                break;
            case 'M':
            case 'N':
            case 'O':
                cout << 6;
                break;
            case 'P':
            case 'Q':
            case 'R':
            case 'S':
                cout << 7;
                break;
            case 'T':
            case 'U':
            case 'V':
                cout << 8;
                break;
            case 'W':
            case 'X':
            case 'Y':
            case 'Z':
                cout << 9;
                break;
            default:
                cout << "[invalid]";
            }

            if (noOfLetters == 3)
            {
                cout << '-';
            }
            else if (noOfLetters > 7)
            {
                cin.ignore();
            }
            cin >> letter;

            //noOfLetters++;
        }
        cout << endl;
        cin.ignore(100, '\n');

        cout << "\nTo process another telephone number, enter Y/y \n"
             << "Enter any other letter to terminate the program: ";
        cin >> response;
       }
}

如果我输入超过 7 个字符，一切正常。问题是，如果我正好输入 7 个字符，它就会把自己弄乱。

假设我输入“honk honk”(不包括空格，长度为 8 个字符)并运行程序。输出如下:

Enter Y/y to convert a telephone number form letters to digits. Enter any other letter to terminate the program: y

Enter a telephone number using letters: honk honk The corresponding telephone number is: 466-5466

To process another telephone number, enter Y/y Enter any other letter to terminate the program: n Press any key to continue . . .

所以，这很好用。但是如果我输入“七十”(恰好 7 个字符)，程序会强制我在继续之前输入另一个字符(不包括空格和换行符)，如下所示:

Enter Y/y to convert a telephone number form letters to digits. Enter any other letter to terminate the program: y

Enter a telephone number using letters: seventy The corresponding telephone number is: 738-3689 h

To process another telephone number, enter Y/y Enter any other letter to terminate the program: n Press any key to continue . . .

我想修改程序，以便我能够在程序中输入 7 个字符的短语并使其正确运行而不会出现任何错误，例如当我输入任何长度为 8 个字符或更多的字符时。

我已经尝试了所有我能做的，但没有运气。有人可以提出建议来帮助我修复此错误吗？

Answer 1

你的逻辑流程有点困惑。请注意，当 noOfLetters 为七时，您仍会在 while 循环结束时要求另一个字母。考虑在循环的顶部读取。如果这样更容易，请缓冲循环中的输出。

cout << "Enter a telephone number using letters: " << flush;
// Flush may be necessary if output is line-buffered  ^

noOfLetters = 0;

// This is a buffer where we will be storing the output phone number so that we
// don't have to deal with mixing input with output.
//
// You will need to "#include <sstream>".
stringstream phoneNumber;

// Reading becomes part of the loop condition; if the end of the input is reached,
// we want the loop to terminate.  "cin >> letter" will evaluate to false in
// boolean context if reading failed.
//
// Note that && short-circuits; if the left side is false then the right side is
// not even evaluated.  So when noOfLetters == 7, the loop terminates without
// reading another character.
while (noOfLetters != 7 && cin >> letter) {
    ++noOfLetters;

    // Your switch block goes here. Replace "cout" with "phoneNumber".

    if (noOfLetters == 3) phoneNumber << '-';
}

cout << "The corresponding telephone number is: " << phoneNumber.rdbuf() << endl;