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So I have a question about bool variables.

This is a program which checks whether the due is payed on time and if it is not, it is multiplied by 1.10.

#include <iostream> using namespace std; int main() { float Dues; cout<<"Enter ammount: \n"; cin>>Dues; cout<<"On time? (y/n)"; char yn; cin>>yn; bool Overdue = yn !="y"; //TRUE (1) if it is late, FALSE (0) if it is on time float AmountDue; AmountDue = Overdue ? Dues*1.10 : Dues; cout<<"Ammount due: "; cout<<<<AmountDue; return 0; }

I don't undestand the logic of the bool

We have

bool Overdue = yn !="y";

Now this is my understaning of the logic of the bool and it is NOT right

If "n" is entered => N is NOT Y which is CORRECT therefore the bool is true => 1

If "y" is entered => Y is NOT Y which is WRONG, therefore fasle => 0

But it is actually the other way around and I can't explain it logically to myself. On what logic is based bool Overdue = yn !="y"; ?

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  • 1
    bool Overdue = yn !='y'; Commented Dec 14, 2012 at 13:32
  • 1
    That's the worst spell of weather I've ever seen! :) (try whether) Commented Dec 14, 2012 at 13:46
  • lol I am writing completely different word :D Commented Dec 14, 2012 at 13:56

2 Answers 2

Reset to default
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In addition to jrok's answer the problem you are encountering is that you assume that lowercase and uppercase characters are the same thing. They are NOT. 'y' and 'Y' are two different characters. Same thing for 'n' and 'N'.

You write:

If "n" is entered => N is NOT Y which is CORRECT therefore the bool is true => 1

No. 'n' is not 'y'.

If "y" is entered => Y is NOT Y which is WRONG, therefore fasle => 0

It's CORRECT. 'y' is NOT 'Y'.

Try this instead:

bool Overdue = (yn != 'n') && (yn != 'N');
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1 Comment

Ah, +1, it didn't occur to me this could confuse OP.
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The reason for unexpected behaviour is that you're comparing a char with a string literal "y". String literals are of type const char[n] where n is the length of the literal including the terminating NUL character.

Compare with a character literal instead:

yn != 'y'

When you say this: yn != "y", the char gets promoted to int and the string literal decays to const char*. How is this supposed to behave is unspecified by the standard.

The result of the expression gets stored in bool Overdue. When yn holds 'n', the expression is true ('n' is indeed different than 'y') so true will be stored, and vice versa).

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12 Comments

I don't see any 1s or 0s in your code. do you mean true and false?
After you change to char literal, your program behaves correctly, AFAICT.
@user1888353 Umm, with the type correction suggested by jrok (that is, as the question stands now), it works as it should (entering 'n' makes overdue have value true).
@user1888353 In C++, bool is an actual type and true and false are its values. While they convert to 1 and 0 when converted to a number, it's better to think about them as boolean values, not as numbers.
It's not. When you enter 'y', the expression yn != 'y' becomes 'y' != 'y', which clearly isn't true - the result of expression is false and that's what ends up stored in Overdue. Try again and see for yourself.
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