Logical OR and ternary operator evaluation order

Question

Here, operators with the highest precedence appear at the top of the table, those with the lowest appear at the bottom. Within an expression, higher precedence operators are evaluated first.

Logical OR || Left to right

Conditional ?: Right to left

Example:

if ( true || false ? false : true )
{
     Console.WriteLine(String.Format("WHY?"));
}

How I see it should be: true || (false ? false : true)

How I think it work in my case: (true || false) ? false : true

Why the condition will never be fulfilled if the conditional OR has a higher priority over the ternary operator?

UPD1: Look, the first and third condition are marked unreachable

UPD2: The logical OR should not even check the conditions following it, if it has already found one true

Answer 1

Operator precedence is all about solving ambiguities. Let's say for example we have two operators op1 and op2 and the following expression:

x = a op1 b op2 c

We could resolve this logically as

tmp = a op1 b
x = tmp op2 c

But it would also be possible to evaluate op2 first, like so:

tmp = b op2 c
x = a op1 tmp

This is ambigous. To solve the ambiguity we could either demand parentheses or have to give one of the operators higher precedence. This is basically the same what we use in mathematics with an expression like 1 + 2 * 3 where the multiplication has higher precedence and has to be calculated first.

Wat does that mean for your example? As you've already written, || has a higher precedence than ?:. This means that

true || false

will be evaluated first and always results in true. So your ternary effectively becomes

true ? false : true

and will always result in false.

How does the compiler reach that decision? Let's have a (simplified) look at the steps:

1. initially the compiler sees true which is already a value.
2. the compiler sees || an knows now that there has to follow another value or evaluated expression.
3. the compiler then sees false. This would be a value but since it could be part of another expression it's not done yet.
4. next, the compiler encounters ? which it identifies this as the ternary operator
5. since the ternary has a lower precedence than || it has to treat the entire expression true || false as the first operand of the ternary.

As to why the ternary operator has a lower precedence than || I can only speculate that it is for consistency. The ternary is not only the only operator with three operands but also only the second and the third need to have the same datatype whereas the first is always a bool. The lower precedence ensures that these two assignments are semantically the same:

bool b = condition1 || condition2 ? true : false;
int i = condition1 || condition2 ? 23 : 42;