CODEWITHSUNDEEP
c++typescomparison
pighead10
pighead10

Reputation: 4305

Comparing data types

int item;
cin >> item;

That's in my code, but I want the user to be able to type integers or strings. This is basically what I want to do:

if(item.data_type() == string){
  //stuff
}

Is this possible?

Upvotes: 2

Views: 3813

Answers (4)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361802

What you're doing is not value C++ code. It wouldn't compile!

Your problem is:

That's in my code, but I want the user to be able to type integers or strings

Then do this:

std::string input;
cin >> input;
int intValue;
std::string strValue;
bool isInt=false;
try 
{   
    intValue = boost::lexical_cast<int>(input);
    isInt = true;
}
catch(...) { strValue = input; }

if ( isInt) 
{
   //user input was int, so use intValue;
}
else
{
   //user input was string, so use strValue;
}

Upvotes: 0

SoapBox
SoapBox

Reputation: 20609

You can't do exactly that, but with a little more work you can do something similar. The following code works if you have the Boost libraries installed. It can be done without boost, but its tedious to do so.

#include <boost/lexical_cast.hpp>

main() {
    std::string val;
    std::cout << "Value: " << std::endl;
    std::cin >> val;
    try {
        int i = boost::lexical_cast<int>(val);
        std::cout << "It's an integer: " << i << std::endl;
    }
    catch (boost::bad_lexical_cast &blc) {
        std::cout << "It's not an integer" << std::endl;
    }
}

Upvotes: 2

Eugene Burtsev
Eugene Burtsev

Reputation: 1475

Try to read this articles:

Use RTTI for Dynamic Type Identification

C++ Programming/RTTI

Run-time type information

Upvotes: 0

Mark.Ablov
Mark.Ablov

Reputation: 887

no, but you can input string and then convert it to integer, if it is integer.

Upvotes: 1

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