CODEWITHSUNDEEP
c++boolean-logic
user1888353
user1888353

Reputation: 47

Bool variable. Logic of the check

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"; ?

Upvotes: 1

Views: 142

Answers (2)

Nik Bougalis
Nik Bougalis

Reputation: 10613

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');

Upvotes: 2

jrok
jrok

Reputation: 55395

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).

Upvotes: 2

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