Why does cin.clear() fix an infinite loop caused by bad input to cin?

问题

I've written a switch statement, and created a default which would simply say the user picked a bad option and repeat the input. I wanted to make sure that if there was an issue, it would clear the buffer first, so I used cin.sync(), but entering an 'a' for the input still caused an infinite loop. I added cin.clear() to clear the flags which gave me working code ... but my confusion is why it worked. Does cin.sync() not work if there is a fail flag?

the statement follows, truncated for brevity:

while(exitAllow == false)
{
    cout<<"Your options are:\n";
    cout<<"1: Display Inventory\n";
    /*truncated*/
    cout<<"5: Exit\n";
    cout<<"What would you like to do? ";

    cin.clear();   //HERE IS MY CONFUSION//
    cin.sync();
    cin>>action;
    switch(action)
    {
        case 1:
                    /*truncated*/
        case 5:
            exitAllow = true;
            break;
        default:
            cout<<"\ninvalid entry!\n";
            break;
    }
}

回答1:

Once you try to read and it fails, the stream is marked to be in a fail state, and all subsequent attempts to read will fail automatically until you clear the fail flag, which is done with the clear() function.

You should check the status of the stream after each read:

if (cin >> var) {
   // do something sensible
} else {
   cin.clear();
}

Otherwise the value stored in var after the last successful read will be considered to be the current input for your algorithm.

来源：https://stackoverflow.com/questions/13715131/why-does-cin-clear-fix-an-infinite-loop-caused-by-bad-input-to-cin