C++ cin.get(); functioning oddly?

Question

I'm new to C++. I was messing around creating a simple program that took the two numbers you inputted and displayed the answer. The only problem was that it closed too fast. I decided to do what I usually do and and a cin.get(); and it usually solves the problem. The weird thing is, this time it didn't. I had to put two cin.get(); statements. I'm curious to why it requires me to put two for it to stay open instead of the usual one. Here is my code:

int a;
    cin >> a;
    int b;
    cin >> b;
    cout << a + b;
    cin.get();
    cin.get();

Tahlil · Accepted Answer · 2014-05-23T04:57:47.740

4

At the last input (b value) you are also giving a new line as input (when you press enter). So the first cin.get() is taking that new line as another input. And the last one then keeps your console open.

Adding cin.ignore(); after cin>>b; should solve the problem.

edited May 23 '14 at 04:57

answered May 21 '14 at 01:29

Tahlil

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How would I fix the problem? – Meme Machine May 21 '14 at 01:36
The Rolls-Royce version is `cin.ignore(numeric_limits::max(), '\n');` which means to discard all of the rest of the input (if any) after the `b` was read. – M.M May 21 '14 at 02:12
won't Just adding `cin.ignore()` after `cin>>b` work? – Tahlil May 21 '14 at 02:40

score 1 · Answer 2 · 2015-08-31T08:20:15.313

1

Use cin.ignore(); instead. That should fix your problem.

edited Aug 31 '15 at 08:20

answered Jan 21 '15 at 18:57

C++ cin.get(); functioning oddly?

2 Answers2