如何爲開關盒提供強大的用戶輸入？

Question

我想爲開關盒做一個完全防錯輸入。如果用戶輸入錯誤的數字，字母或長串數字或字母（這是我之前有過錯誤的地方），它不會失敗。

防止錯誤，如果用戶輸入，例如。 "asdghdk3"我曾嘗試使用數組，因此它會檢查每個字母，直到它找到一個數字。

我試圖把它變成一個整數的開關情況。 可悲的是我的代碼不起作用。有沒有人有任何建議或改進？ 謝謝。

#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; 
} 

編輯： 我想它只能接受1，2或3，其他一切返回不正確的輸入，請重試。

using namespace std;

int main() 
{ 
    string line; 
    getline(cin, line); 
    istringstream choice(line); 
    int number; 
    choice >> number; 

    if (choice) 
    { 
     if (choice == 1) 
     { 
      cout << "you chose option 1\n"; 
      //code..... 
     } 
     else if (choice == 2) 
     { 
      cout<< "you chose option 2\n"; 
      //code..... 
     } 
     else if (choice == 3) 
     { 
      cout<< "you chose option 3\n"; 
      //code...... 
     } 
    } 
    else 
    { 
     cout << "input does not start with a number or is too big for an int\n"; 
    } 
return 0; 
} 

Answer 1

閱讀從std::cin一整行與std::getline一個std::string，那麼該行轉換爲整數與std::istringstream。最後，在轉換之後，檢查字符串流中是否還有字符。這裏有一個完整的例子：

#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 number\n"; 
     } 
     else 
     { 
      std::cout << "input ok\n"; 
     } 
    } 
    else 
    { 
     std::cerr << "inut does not start with a number or is too big for an int\n"; 
    } 
} 

Answer 2

您應該使用std::cin並檢查其狀態：

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 
}