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Home/ Questions/Q 6987217
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T18:54:30+00:00 2026-05-27T18:54:30+00:00

Here is my simple code: #include <iostream> int main() { int foo; std::cin.rdbuf(std::cout.rdbuf()); std::cin>>foo;

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Here is my simple code:

#include <iostream>
int main()
{
    int foo;
    std::cin.rdbuf(std::cout.rdbuf());
    std::cin>>foo; // what'll happen at this line? whatever I'll input will go to cout's buffer then to foo , right?
}

What I was thinking that above code will set cin‘s buffer to cout‘s buffer so when I will input some number it’ll will be outputted also. I guess I’m confused with my own program. Can anyone tell me what’s going on in the program?

Also, if I add one more line at end : std::cout<<foo; , then it prints random number which means foo never gets input. So what’s happening overall?

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  1. Editorial Team
    Editorial Team
    2026-05-27T18:54:31+00:00Added an answer on May 27, 2026 at 6:54 pm

    The stream is responsible for the formatting and delegates the IO to the streambuf (which thus do more than buffering, it executes the IO as well).

    So with std::cin.rdbuf(std::cout.rdbuf()) you are asking to cin to do its input using the streambuf of cout, which probably is not ready for doing input. So std::cin>>foo will fail.

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