c++ validate input value before add it to array

Question

My target is to validate input value before add it into array. Current code used:

int main()
{
    int temp;
    int arr[5];
    for(int i = 0; i < 5; i++)
    {
        // validate here
        cin >> arr[i];
    }
    return 0;
}

and my validation method:

int validateInput(string prompt)
{
    int val;
    while (true)
    {
        cin.clear();
        cin.sync();
        cout << prompt;
        cin >> val;
        if (cin.good() && val >= -50 && val <= 50)
        {
            break;
        }
        else
            cin.clear();
        cout << "Invalid input! number must be between -50 and 50" << endl;
    }
    return val;
}

How is that possible?

You have a function that reads some input and "validates" it, what is the problem with it? — Some programmer dude
– Some programmer dude, Commented Mar 22, 2017 at 15:14
With C++, you could build something that enables you to write cin >> Foo(arr[i], -50, 50). Google "overloading istream >>". Alas, the boring simple way is probably more practical. — Bathsheba
– Bathsheba, Commented Mar 22, 2017 at 15:16
Prefer to use std::vector. Arrays have the possibility of buffer overflow, and are difficult to pass to functions (compared to vectors). — Thomas Matthews
– Thomas Matthews, Commented Mar 22, 2017 at 15:31

Vittorio Romeo · Accepted Answer · 2017-03-22 16:01:05Z

2

Your validateInput should just deal with validation: it should answer "is x valid or not valid?"

bool validateInput(int x)
{
    return val >= -50 && val <= 50;
}

When reading from stdin, use validateInput and branch accordingly:

for(int i = 0; i < 5; i++)
{
    int temp;
    cin >> temp;

    if(cin.good() && validateInput(temp))
    {
        arr[i] = temp;
    }
    else
    {
        cout << "Invalid input! number must be between -50 and 50" << endl;
        // handle invalid input
    }
}

If you want to further abstract the idea of "only reading valid numbers from std::cin", you can use an higher-order function:

template <typename TFValid, typename TFInvalid, typename TFInvalidIO>
decltype(auto) processInput(TFValid&& f_valid, TFInvalid&& f_invalid, TFInvalidIO&& f_invalid_io)
{
     int temp;
     cin >> temp;

     if(!cin.good()) 
     {
         // Invalid IO.
         return std::forward<TFInvalidIO>(f_invalid_io)();
     }

     if(validateInput(temp))
     {
         // Valid IO and datum.
         return std::forward<TFValid>(f_valid)(temp);
     }

     // Valid IO, but invalid datum.
     return std::forward<TFInvalid>(f_invalid)(temp);
}

Usage:

for(int i = 0; i < 5; i++)
{
    processInput([&](int x){ arr[i] = x; },
                 [](int x){ cout << x << " is invalid"; },
                 []{ cout << "Error reading from cin"; });
}

If you want more genericity you can also pass validateInput and the type of input as an additional parameters.

edited Mar 22, 2017 at 16:01

answered Mar 22, 2017 at 15:15

Vittorio Romeo

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6 Comments

Some programmer dude Over a year ago

You might want to add the cin >> temp as part of the condition in the if statement.

IntoTheDeep Over a year ago

No no, I have validation method, which I want to combine in main

IntoTheDeep Over a year ago

In your answer if input is not valid, it will not prompt again. In my method it prompts till valid

vsoftco Over a year ago

You most likely want cin.good() && validateInput(temp), i.e. && instead of logical bit-and &. For bool the only difference will be in that && does short-cicuit evaluation, but for non-bools, & is very different from &&.

Vittorio Romeo Over a year ago

@vsoftco: that was a typo :)

|

Harry · Accepted Answer · 2017-03-22 15:23:17Z

0

Vittorio's answer above is spot on. For completeness, if you want exactly 5 elements:

#include <string>
#include <iostream>

using namespace std;

bool validateInput(int inValue)
{
  return inValue >= -50 && inValue <= 50;
}

int main()
{
  int _arr[5];
  int _currentIdx = 0;
  int _tmp;
  while (_currentIdx < 5)
  {
    cin >> _tmp;
    if (validateInput(_tmp))
    {
      _arr[_currentIdx] = _tmp;
      _currentIdx++;
    }
    else
    {
      cout << "Invalid input! number must be between -50 and 50" << endl;
    }
  }
}

answered Mar 22, 2017 at 15:23

Harry

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2 Comments

IntoTheDeep Over a year ago

In your answer if input is not valid, it will not prompt again. In my method it prompts till valid

Harry Over a year ago

@TeodorKolev How so? If input is invalid _currentIdx is not increased and it will prompt again for the given array index.

IntoTheDeep · Accepted Answer · 2017-03-22 17:52:52Z

0

replace

cin >> temp;

with

arr[i] = validateInput("some str");

edited Mar 22, 2017 at 17:52

answered Mar 22, 2017 at 15:42

IntoTheDeep

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1 Comment

Toby Speight Over a year ago

There's no temp in the question. Are you sure this is an answer?

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