如果我为c ++中的int提供double值，那么存储在输入缓冲区中的内容是什么？

Question

My code is :

int main()
{
    int x,y,z;
    cout<<"Please input first integer : ";
    x = getInt();
    cin>>z>>y;
    cout<<x <<"\n"<<z<<"\n"<<y;

    return 0;
}

when I provide input such as

Please input first integer : 34.5
34
0
-2
Process returned 0 (0x0)   execution time : 3.431 s
Press any key to continue.

What's happening over here??

Answer 1

When trying to read an integer (using >> ), the input 34.5 is no different from the strings 34abcd or 34 42 .

The operator>> reads characters that could be part of an integer, 3 and 4 , and then stops (leaving the rest in the input buffer).

When continuing with cin >> z the buffer still contains .5 . As the . cannot be part of an integer, the input fails and z is set to 0.

After that, the stream is in a failed state and the input to y isn't even attempted. The stream state has to be cleared if we want to try more input.

---

The result is complicated by the fact that y is still uninitialized when the values are displayed. The -2 is one possible effect of this undefined behavior.

Answer 2

If extraction fails (eg if a letter was entered where a digit is expected), value is left unmodified and failbit is set.

As results, subsequent calls to << are failed (failbit is set).