Do While Loop Question

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.

P.S.: What game is this? Sounds like Crazy Taxi. :-)

user541686
– user541686

2011-05-23 23:28:07 +00:00
Commented May 23, 2011 at 23:28 — user541686
– user541686, Commented May 23, 2011 at 23:28

YXD · Accepted Answer · 2011-05-23 23:28:12Z

16

Learn De Morgan's laws

(not A) or (not B)

is not the same as

not (A or B).

answered May 23, 2011 at 23:28

YXD

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Comments

Oliver Charlesworth · Accepted Answer · 2011-05-23 23:29:20Z

3

(ch != 'q' || ch != 'Q') is always true: "ch is not equal to 'q' or ch is not equal to 'Q'".

answered May 23, 2011 at 23:29

Oliver Charlesworth

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Comments

JaredPar · Accepted Answer · 2011-05-23 23:29:26Z

2

The problem is your boolean logic is off and the two while conditions are not the same.

Top: Character is not 'q' or is not 'Q'
Bottom: Character is not ('q' or 'Q')

The Top will return true for every single character possible. The bottom will return true for every character except 'q' and 'Q'

answered May 23, 2011 at 23:29

JaredPar

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Comments

jwueller · Accepted Answer · 2011-05-23 23:28:51Z

0

I think you want this in your first example:

ch !='q' && ch != 'Q'

You want that the input is not q AND not Q.

answered May 23, 2011 at 23:28

jwueller

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Adam Byrtek · Accepted Answer · 2011-05-23 23:29:25Z

0

!(ch=='q' || ch=='Q') is equivalent to ch!='q' && ch!='Q'. See also De Morgan's laws.

answered May 23, 2011 at 23:29

Adam Byrtek

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Comments

SoapBox · Accepted Answer · 2011-05-23 23:32:59Z

0

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.

answered May 23, 2011 at 23:32

SoapBox

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Comments

MikeTeeVee · Accepted Answer · 2011-05-23 23:39:55Z

0

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');

answered May 23, 2011 at 23:39

MikeTeeVee

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1 Comment

user705185 Over a year ago

Thanks, I'm probably going to use a similar function to ToUpper() in the future.

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