Exam ques. on how booleans are handled in cout

Question

I'm studying for upcoming exams and came across this past exam question which doesn't make sense to me.

Consider the following main function:

int main()
{
    int x = 0;
    cout << "x = " << x << ", (0 < x < 10) = " << (0 < x < 10) << endl;
    int x = 5;
    cout << "x = " << x << ", (0 < x < 10) = " << (0 < x < 10) << endl;
    int x = 10;
    cout << "x = " << x << ", (0 < x < 10) = " << (0 < x < 10) << endl;

    return 0;
}

When executed the program prints the following:

x = 0, (0 < x < 10) = 1
x = 5, (0 < x < 10) = 1
x = 10, (0 < x < 10) = 1

Explain exactly what has happened.

That's the question. As far as I know, the last line of output should be "x = 10, (0 < x < 10) = 0". What am I missing?

Answer 1

What do you expect 0 < x < 10 to mean?

It doesn't check whether x is between 0 and 10, if that's what you thought.

< is a binary operator, which follows operator evaluation rules (precedence and associativity).

So 0 < x < 10 actually means (0 < x) < 10. You'll need two checks to get the result you want (left to you).

Answer 2

0 < x < 10 does not do what you think it does. It does ( 0 < x ) < 10 which is not what you want. The result of 0 < x ( which will be true or false ) is then checked with < 10 which will also give result as true ( numerically equal to 1 ) or false ( numerically equal to 0 ).

You need

( 0 < x ) && ( x < 10 )

to check whether x in between them.

So, your first cout with x=0 is same as ( 0 < 0 ) which gives false and then 0 < 10 ( false is numerically 0 ), and hence the result is 1.

Similarly, your second cout at x=5 first gives result false and then with 0 < 10 , it gives true.

And finally, your last cout at x=10 first results true, and again true, so result is 1.