How to check if input is numeric in C++

Question

I want to create a program that takes in integer input from the user and then terminates when the user doesn't enter anything at all (ie, just presses enter). However, I'm having trouble validating the input (making sure that the user is inputting integers, not strings. atoi() won't work, since the integer inputs can be more than one digit.

What is the best way of validating this input? I tried something like the following, but I'm not sure how to complete it:

char input

while( cin>>input != '\n')
{
     //some way to check if input is a valid number
     while(!inputIsNumeric)
     {
         cin>>input;
     }
}

`input` is a pointer. Why are you taking input to it using `cin` ? I am not sure whether an address can be given that way but address locations are represented in **hexa-decimal** system and there chances of being alpha-numeric. — Mahesh, Apr 13 '11 at 20:25
Gah, I was recreating this code from memory (not copying and pasting). Let me try and fix that. — chimeracoder, Apr 13 '11 at 20:26
your while condition won't compile. Perhaps you should go with `for(cin>>input; input != '\n'; cin >>input)` — Tim, Apr 13 '11 at 20:45

greyfade · Accepted Answer · 2015-04-15T06:46:45.653

52

When cin gets input it can't use, it sets failbit:

int n;
cin >> n;
if(!cin) // or if(cin.fail())
{
    // user didn't input a number
    cin.clear(); // reset failbit
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); //skip bad input
    // next, request user reinput
}

When cin's failbit is set, use cin.clear() to reset the state of the stream, then cin.ignore() to expunge the remaining input, and then request that the user re-input. The stream will misbehave so long as the failure state is set and the stream contains bad input.

edited Apr 15 '15 at 06:46

answered Apr 13 '11 at 21:11

greyfade

24,948
7
64
80

2

The problem with that (against my reading of the Q) is that it doesn't give up after one press of ; rather - `cin >> n` will keep running until success, error or end-of-file. – Tony Delroy Apr 14 '11 at 02:23
@Tony: This is true, but it also doesn't advance the read pointer until it has a successful read. I've edited my post. – greyfade Apr 14 '11 at 03:42
3

Don't forget to `.reset()` the stream. – Lightness Races in Orbit May 20 '11 at 17:11
Oddly this does not work on G++-47 compiler (Mint 16). No idea why. – Ace Caserya Apr 15 '15 at 01:19
@AlvinCaseria: Could you clarify what you mean by "does not work"? – greyfade Apr 15 '15 at 04:32
My bad. I didn't notice the clear and ignore, which was my problem. I suggest code be edited to include them both. – Ace Caserya Apr 15 '15 at 05:18
compiling in Xcode this works perfectly but if using g++ you have to `#include ` in order for it to work – MiladiuM Oct 26 '16 at 16:47

score 23 · Answer 2 · edited Aug 28 '13 at 18:22

23

Check out std::isdigit() function.

edited Aug 28 '13 at 18:22

answered Apr 13 '11 at 20:27

Christian

7,433
4
36
61

HappyTran · Answer 3 · 2015-06-03T02:34:07.617

8

The problem with the usage of

cin>>number_variable;

is that when you input 123abc value, it will pass and your variable will contain 123.

You can use regex, something like this

double inputNumber()
{
    string str;
    regex regex_pattern("-?[0-9]+.?[0-9]+");
    do
    {
        cout << "Input a positive number: ";
        cin >> str;
    }while(!regex_match(str,regex_pattern));

    return stod(str);
}

Or you can change the regex_pattern to validate anything that you would like.

edited Jun 03 '15 at 02:34

answered Jun 02 '15 at 08:05

HappyTran

503
6
9

this is great idea in c++ 11 – Albert Hidalgo Oct 03 '18 at 01:45

score 3 · Answer 4 · answered Apr 13 '11 at 20:50

I find myself using boost::lexical_cast for this sort of thing all the time these days. Example:

std::string input;
std::getline(std::cin,input);
int input_value;
try {
  input_value=boost::lexical_cast<int>(input));
} catch(boost::bad_lexical_cast &) {
  // Deal with bad input here
}

The pattern works just as well for your own classes too, provided they meet some simple requirements (streamability in the necessary direction, and default and copy constructors).

Valentin Galea · Answer 5 · 2011-12-09T14:31:59.823

2

Why not just using scanf("%i") and check its return?

edited Dec 09 '11 at 14:31

answered Apr 13 '11 at 21:06

Valentin Galea

1,084
8
18

score 0 · Answer 6 · answered Jan 23 '14 at 18:18

If you already have the string, you can use this function:

bool isNumber( const string& s )
{
  bool hitDecimal=0;
  for( char c : s )
  {
    if( c=='.' && !hitDecimal ) // 2 '.' in string mean invalid
      hitDecimal=1; // first hit here, we forgive and skip
    else if( !isdigit( c ) ) 
      return 0 ; // not ., not 
  }
  return 1 ;
}

score 0 · Answer 7 · answered Apr 13 '11 at 21:18

I guess ctype.h is the header file that you need to look at. it has numerous functions for handling digits as well as characters. isdigit or iswdigit is something that will help you in this case.

Here is a reference: http://docs.embarcadero.com/products/rad_studio/delphiAndcpp2009/HelpUpdate2/EN/html/devwin32/isdigit_xml.html

score 0 · Answer 8 · answered Feb 06 '22 at 00:13

Something that can be done is read text input and then try to convert that to a number:

try 
{
    std::string str;
    std::getline(std::cin, str)
    age = std::stoi(str)
}
catch(...)
{
    // ...
}

Then you can handle the errors by catching the exceptions.

https://en.cppreference.com/w/cpp/string/basic_string/stol

How to check if input is numeric in C++

8 Answers8

Linked

Related