How do I check if input is an integer/string?

Question

My problem here is I don't know how to insert a rule wherein if a user inputted a number on the string, it will cout a warning saying it's not valid, same with if a user inputted a string/char on the grades. How? I've been trying it but the formula won't work.

int x, cstotal = 100, extotal = 150;

double scorecs, exscore, labtotala, labtotalb, total;

string mystr = "";

cout << "Compute for: " << "\n" << "1. Laboratory Grade " << "\n" << "2. Lecture Grade" << "\n" << "3. Exit" << "\n";
cout << "Please enter a number: ";
cin >> x;
switch (x) {
case 1:

    cout << "Compute for laboratory grade." << "\n";
    cout << "Enter Student Name: ";
    cin >> mystr;


    cout << "Good day, " << mystr << " . Please provide the following grades: " << "\n";
    cout << "CS Score: ";
    cin >> scorecs;

    cout << "Exam Score: ";
    cin >> exscore;


    labtotala = scorecs / cstotal * 0.6;
    labtotalb = exscore / extotal * 0.4;
    total = labtotala + labtotalb;
    cout << "Your Laboratory Grade is " << total * 100 << "\n";
    system("pause");
    break;
case 2:
    cout << "Compute for lecture grade." << "\n";
    cout << "Enter Student Name: ";
    cin >> mystr;
    cout << "Good day, " << mystr << " . Please provide the following grades: " << "\n";
    cout << "CS Score: ";
    cin >> scorecs;
    cout << "Exam Score: ";

    cin >> exscore;
    labtotala = scorecs / cstotal * 0.7;
    labtotalb = exscore / extotal * 0.3;
    total = labtotala + labtotalb;
    cout << "Your Lecture Grade is " << total * 100 << "\n";
    system("pause");
    break;

What part of the rule is giving you trouble? How to implement the detection, control flow, branching? If it's just a matter of detecting if a string is a number, you could loop through each character and call isdigit. — RonaldBarzell
– RonaldBarzell, Commented Nov 18, 2012 at 14:05
you need to be more specific as to what you are looking for. Post less code if possible. — Dylan L.
– Dylan L., Commented Feb 9, 2016 at 1:55

Aamir · Accepted Answer · 2012-11-18 14:08:18Z

14

cin sets a failbit when it gets input of an invalid type.

int x;
cin >> x;

if (!cin) {
    // input was not an integer
}

You can also use cin.fail() to check if the input was valid:

if (cin.fail()) {
    // input was not valid
}

answered Nov 18, 2012 at 14:08

Aamir

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Comments

Some programmer dude · Accepted Answer · 2016-02-09 07:18:56Z

4

How about something like this:

std::string str;
std::cin >> str;

if (std::find_if(str.begin(), str.end(), std::isdigit) != str.end())
{
    std::cout << "No digits allowed in name\n";
}

The above code loops through the whole string, calling std::isdigit for each character. If the std::isdigit function returns true for any character, meaning it's a digit, then std::find_if returns an iterator to that place in the string where it was found. If no digits were found then the end iterator is returned. This way we can see if there was any digits in the string or not.

The C++11 standard also introduced new algorithm functions that can be used, but which basically does the above. The one that could be used instead is std::any_of:

if (std::any_of(str.begin(), str.end(), std::isdigit))
{
    std::cout << "No digits allowed in name\n";
}

edited Feb 9, 2016 at 7:18

answered Nov 18, 2012 at 14:08

Some programmer dude

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3 Comments

Dylan L. Over a year ago

So will this check whether or not an int is inside of the input? if so will you please explain your code?

Some programmer dude Over a year ago

@DylanLittle Updates my answer trying to explain what it does, and with links to references. Also added a "better" solution when using C++11 or later.

Dylan L. Over a year ago

ahhh so much better. I can see what's going on now.

poorva · Accepted Answer · 2013-12-01 11:27:04Z

3

    cout << "\n Enter number : ";
    cin >> ch;
    while (!cin) {
        cout << "\n ERROR, enter a number" ;
        cin.clear();
        cin.ignore(256,'\n');
        cin >> ch;
    }

answered Dec 1, 2013 at 11:27

poorva

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Comments

Rahul Tripathi · Accepted Answer · 2012-11-18 14:06:38Z

1

Use the .fail() method of the stream. Something like below:-

   cin >> aString;

  std::stringstream ss;
  ss << aString;
  int n;
  ss >> n;

  if (!ss.fail()) {
   // int;
  } else {
  // not int;
   }

answered Nov 18, 2012 at 14:06

Rahul Tripathi

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Comments

Amir Sasani · Accepted Answer · 2016-05-30 15:24:13Z

1

you could use cin.fail() method! When cin fails it will be true and you could us a while loop to loop until the cin is true:

cin>>d;
while(cin.fail()) {
    cout << "Error: Enter an integer number!"<<endl;
    cin.clear();
    cin.ignore(256,'\n');
    cin >> d;
}

answered May 30, 2016 at 15:24

Amir Sasani

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1 Comment

SHAH MD IMRAN HOSSAIN Over a year ago

It shold be just cin.ignore() in new compiler. cin.ignore(256,'\n') doesn’t work in updated compiler

feras abo-aloun · Accepted Answer · 2020-01-08 01:10:24Z

0

//this program really work in DEV c++
#include <iostream> 
using namespace std; 
int main()
{
    char input;
    cout<<"enter number or value to check"<<endl;
    cin>>input;
for(char i='a';i<='z';i++)
{
    if(input==i)
    {
        cout<<"character"<<endl;
        exit(0);
    }
}
for(int i=0;i<=1000;i++)
{
    if(input==i)
    {
        cout<<"number"<<endl;   
    }
}
}

answered Jan 8, 2020 at 1:10

feras abo-aloun

1

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