When and why do I need to use cin.ignore() in C++?

Raddicus picture Raddicus · Aug 24, 2014 · Viewed 225.9k times · Source

I wrote a very basic program in C++ which asked the user to input a number and then a string. To my surprise, when running the program it never stopped to ask for the string. It just skipped over it. After doing some reading on StackOverflow, I found out that I needed to add a line that said:

cin.ignore(256, '\n');

before the line that gets the string input. Adding that fixed the problem and made the program work. My question is why does C++ need this cin.ignore() line and how can I predict when I will need to use cin.ignore()?

Here is the program I wrote:

#include <iostream>
#include <string>

using namespace std;

int main()
{
    double num;
    string mystr;

    cout << "Please enter a number: " << "\n";
    cin >> num;
    cout << "Your number is: " << num << "\n";
    cin.ignore(256, '\n'); // Why do I need this line?
    cout << "Please enter your name: \n";
    getline (cin, mystr);
    cout << "So your name is " << mystr << "?\n";
    cout << "Have a nice day. \n";

}

Answer

savageWays picture savageWays · Aug 24, 2014

Ignore is exactly what the name implies.

It doesn't "throw away" something you don't need instead, it ignores the amount of characters you specify when you call it, up to the char you specify as a breakpoint.

It works with both input and output buffers.

Essentially, for std::cin statements you use ignore before you do a getline call, because when a user inputs something with std::cin, they hit enter and a '\n' char gets into the cin buffer. Then if you use getline, it gets the newline char instead of the string you want. So you do a std::cin.ignore(1000,'\n') and that should clear the buffer up to the string that you want. (The 1000 is put there to skip over a specific amount of chars before the specified break point, in this case, the \n newline character.)