question about ||

[Rubiks14]

ok i can't seem to figure out what is goin on here. i'm using or in my while statement and an if statement further down and no matter what i type in i can't get the while statement to leave. even if i type n or N. any help would be great

thanks, rubiks14

char stayChoice = 'r';

while((stayChoice != 'n') || (stayChoice != 'N'))
{
	cout << "Are you sure you want to stay at the Inn? (y/n)\n";
	cout << endl;
	cout << ": ";
	cin >> stayChoice;
	cin.ignore();
	cout << endl;

	if((stayChoice == 'y') || (stayChoice == 'Y'))
	{
		if(player.gold >= 50)
		{
			cout << "Your HP/MP have been restored\n";
			SetHealth(player);
			SetMana(player);
			player.gold -= 50;
			cout << "Health: " << player.health << "/" << player.maxHealth << endl;
			cout << "Mana: " << player.mana << "/" << player.maxMana << endl;
		}
		else
		{
			cout << "Sorry you don't have enough gold.\n";
			break;
		}
	}//if(stayChoice == 'y' || stayChoice == 'Y')
}//while(stayChoice != 'n' || stayChoice != 'N')

[SlyMaelstrom]

Look at your loop it proclaims that if the variable doesn't equal 'N' or doesn't equal 'n' it should loop. Which means, in order to exit the loop, the variable would have to equal both 'N' AND 'n' in order to exit. That's impossible. What you want to use in that loop is && not ||

[7stud]

For an OR to be true, only one conditional has to be true. If you enter 'n', will this be true or false:

stayChoice != 'N'

[Rubiks14]

slymaelstrom it works but i thought && was AND and that || was OR. so i am sorta confused here. :P

[7stud]

It's pretty simple, how many of the conditionals have to be true in an OR statement for the whole thing to be true?

[SlyMaelstrom]

Think of it this way... if I were to say to you "Should I punch you or should I slap you", how would you tell me you don't want either of those to happen?

Would you say "I don't want you to punch me OR I don't want you to slap me"
or would you say "I don't want you to punch me AND I don't want you to slap me"?

Note: This is hypothetical, I'm not interested in punching you or slapping you.

[Rubiks14]

just one. but in an AND statement both have to be

good way of puttin it slymaelstrom...and they say violence doesn't solve anything .

[Blackroot]

while((stayChoice != 'n') || (stayChoice != 'N'))

If stayChoice does not equal 'n' or does not equal 'N' then begin. Thats what your loop says.

This looks strange.

char stayChoice = 'r';

Its perfectly legal, but It can cause issues. Try;

char stayChoice[1] = 'r';

That way one space is allocated to the char. Oh, and you need to change all your calls to stayChoice to;

if(stayChoice[0] == 'n' || stayChoice[0] == 'N')
 DoStuff();

That shouldent be an actual problem though.

[7stud]

but in an AND statement both have to be

So if you have this statement:

(stayChoice != 'n') && (stayChoice != 'N')

and stayChoice is equal to 'y', how many of those conditionals are true? And, is the whole thing true or false?

If stay choice is 'n', how many of the conditionals are true? And, is the whole thing true or false?

If stay choice is 'N', how many of the conditionals are true? And, is the whole thing true or false?

Directly translating English language into code doesn't work. You usually have to make a guess at whether it should be AND or OR, but then to make sure you are right, you have to ask yourself the above questions.

[SlyMaelstrom]

Directly translating English language into code doesn't work. You usually have to make a guess at whether it should be AND or OR, but then to make sure you are right, you have to ask yourself the above questions.

"I don't want stayChoice to be 'n' and I don't want stayChoice to be 'N'."

No?

Now, converting to a speakable language could cause problems if you don't say it phonetically. Take, for example:

"I don't want to be punched and I don't want to be slapped" versus
"I don't want to be punched and slapped"

They both mean completely different things, especially when converted into code, but could arguable be understood the same in conversation if one weren't to nitpick.