Thread: First post. Confused about Boolean

O_o

Associativity in C++ can be a pain...

Soma

Like multiplication has a higher precedence than addition, the same goes for boolean AND vs boolean OR.
In your example, the "1 && 0" happens first, giving 0 and then "1 || 0" gives 1, finally "!(1)" gives 0.
The precedence with boolean operators is( from higher to lower ): NOT, AND, OR

You need to consider the precedence and the parentheses. The part in the parens is computed first.
Within the parens, the && has higher precedence, so the first thing that's computed is 1 && 0, which is 0. This yields
!(1 || 0)
Then inside the parens is 1 || 0 which is 1, yielding
!1
And that gives
0

Originally Posted by phantomotap

Associativity in C++ can be a pain...

Are you some kind of paranoid associativity fanboy? I think you mean precedence.

Originally Posted by Dagon's Smile

OK, so 1&&0 is computed first but why is it 0? this is what is confusing me. How does the statement "One and zero" equal false?
Then 1||0 is yields 1. Why? Again I read that part of the tutorial but I don't get it.
I get the last part !1 this is self evident. It's the rest

&& yields 1 if and only if both it's operands are non-zero; otherwise it yields 0.


|| yields 0 if and only if both of it's operands are 0; otherwise it yields 1.

Are you some kind of paranoid associativity fanboy?

O_o

Are you some kind of buffoon?

I'll spell it out for you: I was telling the original poster to search for associativity in C++ because that would take him to "short-circuiting" which would have answered his question.

Because here is a news flash for you: you and GReaper are both wrong!

Code:

!( 1 || 1 && 0 ) 
/*
!( x || y && z ) // labels
*/

Because `x' is true, the expression `y && z' is NEVER EVALUATED AT ALL!

The code example evaluates to false simply because `!(true)' is `false'.

Soma

To clarify the previous post, the logical operators || and && are "short circuiting" or "early out" operators, and the order of evaluation is left to right despite any precedence in the expressions. If the left side of an "short circuiting" operator determines the result, then the right side is not evaluated. In this case (1 || ... ) is going to evaluate to TRUE (1) no matter what the expression is to the right of ||, so the expression to the right of || is not evaluated. Continuing with this example, then !(1 || ...) returns a FALSE (0).

Originally Posted by phantomotap

Are you some kind of buffoon?

I didn't realize you were so fragile.
You sure like name-calling, though.

Hodor, hodor, hodor!

O_o

No Hodor, mommy and daddy aren't getting a divorce; we still love you.

Soma