User input(cin) - Default value

Question

I can\'t figure out how to use a \"default value\" when asking the user for input. I want the user to be able to just press Enter and get the default value. Consider the fol

Answer 1

I'd be tempted to read the line as a string using getline() and then you've (arguably) more control over the conversion process:

int number(20);
string numStr;
cout << "Please give a number [default = " << number << "]: ";
getline(cin, numStr);
number = ( numStr.empty() ) ? number : strtol( numStr.c_str(), NULL, 0);
cout << number << endl;

Answer 2

Use std::getline to read a line of text from std::cin. If the line is empty, use your default value. Otherwise, use a std::istringstream to convert the given string to a number. If this conversion fails, the default value will be used.

Here's a sample program:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int main()
{
    std::cout << "Please give a number [default = 20]: ";

    int number = 20;
    std::string input;
    std::getline( std::cin, input );
    if ( !input.empty() ) {
        std::istringstream stream( input );
        stream >> number;
    }

    std::cout << number;
}

Answer 3

This works as an alternative to the accepted answer. I would say std::getline is a bit on the overkill side.

#include <iostream>

int main() {
    int number = 0;

    if (std::cin.peek() == '\n') { //check if next character is newline
        number = 20; //and assign the default
    } else if (!(std::cin >> number)) { //be sure to handle invalid input
        std::cout << "Invalid input.\n";
        //error handling
    }

    std::cout << "Number: " << number << '\n';    
}

Here's a live sample with three different runs and inputs.

Answer 4

if(!cin)
   cout << "No number was given.";
else
   cout << "Number " << cin << " was given.";