Syncing with stdio is making program i/o non - interactive?

Question

#include <iostream>
using namespace std;

int main()
{   
    ios::sync_with_stdio(0);
    cin.tie(0);

    cout << "Print two numbers: ";
    int x, y;
    cin >> x >> y;
    cout << x << y;
    
    return 0;
}

Input : 23 24 

Output : Print two numbers: 2324

Here before printing the line "Print two numbers: " on console, stdin is waiting for x and y, then prints the entire output on console as given above.

After removing sync lines from above code:

#include <iostream>
using namespace std;

int main()
{   
    // ios::sync_with_stdio(0);
    // cin.tie(0);

    cout << "Print two numbers: ";
    int x, y;
    cin >> x >> y;
    cout << x << y;
    
    return 0;
}

Here first it's printing line "Print two numbers: " on console then stdin stream is waiting for x and y.

I am unable to understand this behaviour.

This behavior is unrelated to syncing with stdio. You changed two lines in your code. — Thomas Sablik, Jul 03 '20 at 18:26
Does this answer your question? [Significance of ios\_base::sync\_with\_stdio(false); cin.tie(NULL);](https://stackoverflow.com/questions/31162367/significance-of-ios-basesync-with-stdiofalse-cin-tienull) — Thomas Sablik, Jul 03 '20 at 18:28

cigien · Answer 1 · 2020-07-03T18:24:28.700

2

By default, cin is tied to cout, so operations on those streams are in the order they are written in the program.

However, doing cin.tie(0) will untie cin from cout, so operations on cout and cin may be interspersed.

Note that all operations on any one stream will still be in the order they are written in the program.

edited Jul 03 '20 at 18:24

answered Jul 03 '20 at 18:15

cigien

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Syncing with stdio is making program i/o non - interactive?

1 Answers1