如何为开关机箱提供可靠的用户输入

How to make a robust user input for a switch case?

本文关键字：用户 输入 开关 机箱      更新时间：2023-10-16

我正在尝试为开关情况进行完全防错的输入。如果用户输入了错误的数字、字母或一长串数字或字母，它不需要失败（这是我以前遇到错误的地方）。

防止用户输入时出错，例如。 "asdghdk3" 我尝试使用数组，因此它会检查每个字母，直到找到一个数字。

然后，我尝试将其转换为开关情况的整数。可悲的是，我的代码将无法正常工作。有人有任何建议或改进吗？谢谢。

#include <iostream>
#include <vector>
#include <cmath>
#include <cstdlib>
using namespace std;
int main()
{
    cout<<"Please choose the method you would like to use to evaluate the potential. Please enter 1,2 or 3:"<<endl;
    cout<<"1. my method. n2. jacobi method. n3. Exit programme. nYou chose: ";
    char choice[20];
    bool k = true;
    int choice2;
    while (k == true){
        fgets(choice, sizeof choice, stdin);
        for(int j=0; j<sizeof(choice); j++){
            if (isdigit(choice[j])==true){  //I want it to check every character until it finds a number.
                choice2 = atoi(choice[j]); //changed name as now integer to be used in switch case.
                k = false;
                break; //so that it breaks out of the for loops because it has found a number
            }
            else{
                continue;
            }
        }
        cout<<"Incorrect input please try again";
    }
    cout<<"nchoice2= "<<choice2<<endl;
    switch ( choice2 ) {
        case 1 :
            // Code
            break;
        case 2:
            // Code
            break;
        case 3:
            //code to exit programme
            break;
        default:
            // Code
            break;
    }
    return 0;
}

编辑：我希望它只接受 1、2 或 3，对于其他所有返回错误输入的内容，请重试。

使用命名空间标准;

int main()
{
    string line;
    getline(cin, line);
    istringstream choice(line);
    int number;
    choice >> number;
    if (choice)
    {
        if (choice == 1)
        {
            cout << "you chose option 1n";
            //code.....
        }
        else if (choice == 2)
        {
            cout<< "you chose option 2n";
            //code.....
        }
        else if (choice == 3)
        {
            cout<< "you chose option 3n";
            //code......
        }
    }
    else
    {
        cout << "input does not start with a number or is too big for an intn";
    }
return 0;
}

用 std::getline 将整行从 std::cin 读到std::string中，然后将该行转换为带有 std::istringstream 的整数。最后，转换后，检查字符串流中是否还剩下字符。下面是一个完整的示例：

#include <iostream>
#include <sstream>
#include <string>
int main()
{
    std::string line;
    std::getline(std::cin, line);
    std::istringstream is(line);
    int number;
    is >> number;
    if (is)
    {
        if (!is.eof())
        {
            std::cerr << "input does not end with a numbern";
        }
        else
        {
            std::cout << "input okn";
        }
    }
    else
    {
        std::cerr << "inut does not start with a number or is too big for an intn";
    }
}
您应该

使用 std::cin 并检查其状态：

int choice = -1;
if (cin >> choice)
{
    // you know user entered a number, check that it's in the correct range
    if (cin.peek() != 'n')
        // there's more input, so probably an error
}
else
{
    // bad input
}
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