c++ - 稍后在程序中使用在 if else 构造内声明的变量会导致未声明的标识符错误

Question

这个程序是不言自明的，所以我不会真正深入了解它的目的。

我现在的主要问题是在第 82、89、95 和 101 行，我在编译时遇到了“arr”和“input”的“Undeclared Identifier”错误。

这是因为我在 if else if 构造中声明了它们吗？如果是这样，有什么办法可以解决这个问题。提前感谢您的帮助!!!!

这是代码

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

template<class T> void selectionSort(T arr[], T num)
{
    int pos_min;
    T temp;

    for (int i = 0; i < num - 1; i++)
    {
        pos_min = i;

        for (int j = i + 1; j < num; j++)
        {
            for (arr[j] < arr[pos_min])
            {
                pos_min = j;
            }
        }


        if (pos_min != i)
        {
            temp = arr[i];
            arr[i] = arr[pos_min];
            arr[pos_min] = temp;
        }
    }
}

int main()
{

    char check = 'C';

    while (toupper(check) != 'Q')
    {
        char dataType;
        int num = 0;

        cout << "What kind of data do you want to sort?" << endl;
        cout << " For integer enter i, for string enter s, for character enter c. ";
        cin >> dataType;

        //User input dataType
        if (toupper(dataType) == 'I')
        {
            int arr[100];
            int input;
            cout << " You've chosen Integer dataType" << endl;
        }
        else if (toupper(dataType) == 'S')
        {
            string arr[100];
            string input;
            cout << " You've chosen String dataType" << endl;
        }
        else if(toupper(dataType) == 'C')
        {
            char arr[100];
            char input;
            cout << " You've chosen Character dataType" << endl;
        }
        else
        {
            cout << "Not a recognizable dataType. Shuting down..." << endl;
            return -1;
        }

        //User input # of num
        cout << "How many num will be sorted? ";
        cin >> num;

        for (int i = 0; i < num; i++)
        {
            cout << "Enter an input of the dataType you selected: ";
            cin >> input;
            arr[i] = input;
        }

        //Display user input
        cout << "The data as you entered it: ";
        for (int i = 0; i < num; i++)
        {
            cout << arr[i];
            cout << " ";
        }
        cout << endl;

        //Sort user input by calling template functon selectionSort
        selectionSort(arr, num);

        //Display sorted user input
        cout << "After sorting your data by calling selectionSort: ";
        for (int i = 0; i < num; i++)
        {
            cout << arr[i];
            cout << " ";
        }

        cout << endl;
        //Query user to quit or continue
    cout << " Would you like to continue? Enter 'Q'. Enter anything else to continue.";
    cin >> check;

    }



    return 0;
}

Answer 1

这是因为您在 if/else block 中声明了它们。一旦 block 完成，这些变量就会离开scope。并且不再可访问。
解决此问题的一种方法是始终将输入作为字符数据读入，然后在事后将其转换为指定的类型。参见 atoi了解如何将 char 转换为 int。