Reputation:

Do While Loop Question

do 
{
  cout << "Car is coming ... " << "[P]ay or [N]ot?" << endl;
  ch=getch();
} while ( ch !='q' || ch != 'Q');

Why will the code on top not work while the code below does? I tried it with parenthesis around each statement in numerous ways and the compiler would pop an error every time until I regrouped them as I did below. I'm just wondering why it does this.

do 
{
  cout << "Car is coming ... " << "[P]ay or [N]ot?" << endl;
  ch=getch();
} while ( !(ch=='q' || ch=='Q') );

I'm using Visual Studio 2008 as my compiler; x86 architecture.

Upvotes: 3

Answers (7)

MikeTeeVee

Reputation: 19392

When inverting all your logic using '!', you did right by reversing the conditional operators "==" to "!=", but you forgot to reverse the logical operators "||" to "&&". Thus, this should be correct:

while (ch!='q' && ch!='Q');

I use C#, so while code above will work, I would have used this instead as it is easier to read:

while (ch.ToUpper() != 'Q');

Upvotes: 0

SoapBox

Reputation: 20609

You've got the logic backwards, that's my negating it works. By DeMirgan's laws, !(ch == 'Q' || ch == 'q') is the same as ch != 'Q' && ch != 'q'.

Since a if it cannot be both little q and big Q at the same time, while (ch != 'Q' || ch != 'q') doesn't make sense because if it is 'Q' then it won't be 'q', and vice versa.

Upvotes: 0