Question

因此，我正在制定飞机座位计划，我希望用户输入数字以选择一行，然后输入字母以选择连续六种选择之一（A到F）。我还希望用户只要输入字母“ C”就可以继续输入座位。我已经编写了大部分代码，但是由于某些原因我无法使其输出，因此在首次初始化并显示座位后，座位不会输出。该程序只是停止。我正在尝试输出结果，然后纠正我在逻辑中犯的任何错误。

//Any unused includes are part of the default code

#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cassert>

using namespace std;

const int rowsz = 5;
const int colsz = 6;

int main()
{
    char seating[rowsz][colsz];
    char y = ' '; //Row
    char x = ' '; //Col
    char taken = 'X'; // This letter is for marking booked seats
    char letter = ' '; // This letter initializes the seat positions.
    char let = ' '; // Choice for a seat
    char choice = 'c'; // This let's the user quit or continue booking seats
    int rownum = 1; // Row number
    for(int row = 1; row <= rowsz; row++)
    {
        letter = 'A';
        cout << rownum;
        rownum++;
        for(int col = 0; col < colsz; col++)
        {
            seating[row][col] = letter;
            letter++;
            cout << seating[row][col];
        }
        cout << endl;
    }
    cout << "Would you like to get a seat? Press C. TO quit, press Q: "; cin >> choice;
    while(toupper(choice) == 'C')
    {
        cout << "Enter a row. Ex: 1,2,3... ";cin >> y;
        cout << "Enter a Letter for a seat: "; cin >> x;

        if(toupper(x) == 'A')
            x = 0;
        if(toupper(x) == 'B')
            x = 1;
        if(toupper(x) == 'C')
            x = 2;
        if(toupper(x) == 'D')
            x = 3;
        if(toupper(x) == 'E')
            x = 4;
        if(toupper(x) == 'F')
            x = 5;

        seating[y][x] = taken;

        for(int row = 1; row <= rowsz; row++)
        {
            for(int col = 0; col < colsz; col++)
            {
                cout << seating[row][col];
            }
            cout << endl;
        }
        cout << "Seat again? Press C to continue to seat and press Q to quit. "; cin >> choice;
    }
    if(toupper(choice) != 'C')
    {
        cout << "Thank you for using this program! " << endl;
    }

    return 0;
}

Answer 1

与提交代码相比，我认为涵盖一些如何调试此类情况的基础知识会更有益。与其一次查看整个程序，不如将其分成几部分，直到找出代码的哪一部分出了问题。例如，让我们看一下本节（提示，提示，提示）：

while(toupper(choice) == 'C')
{
    cout << "Enter a row. Ex: 1,2,3... ";cin >> y;
    cout << "Enter a Letter for a seat: "; cin >> x;

    if(toupper(x) == 'A')
        x = 0;
    if(toupper(x) == 'B')
        x = 1;
    if(toupper(x) == 'C')
        x = 2;
    if(toupper(x) == 'D')
        x = 3;
    if(toupper(x) == 'E')
        x = 4;
    if(toupper(x) == 'F')
        x = 5;

    seating[y][x] = taken;
    for(int row = 1; row <= rowsz; row++)
    {
        for(int col = 0; col < colsz; col++)
        {
            cout << seating[row][col];
        }
        cout << endl;
    }
    cout << "Seat again? Press C to continue to seat and press Q to quit. "; cin >> choice;

尝试在其中插入一些提示，看看要传递给每个变量的内容。如果您花时间用这种方式调试代码，则比我们提供答案要好得多。这些东西一开始总是很费时间，但是如果您有更具体的问题，请随时回来，我们很乐意为您提供帮助！

无法输出可用座位和已占用座位

1 个答案: