In all of the JavaScript operator precedence charts I can find (like this one and this one), the logical AND (&&) has slightly higher precedence to the logical OR (||).
I can't seem to figure out an expression where the result is different than it would be if they had the same precedence. I figure there must be some way for it to matter or they'd be listed as having the same precedence if it didn't.
For example:
0 || 2 && 0 || 3
is 3, but it doesn't matter how I slice that up, it's always going to be 3:
(0 || 2) && 0 || 3
0 || (2 && 0) || 3
(0 || 2 && 0) || 3
0 || 2 && (0 || 3)
0 || (2 && 0 || 3)
If I make that first 0 something else (like 4), the result is always 4 because the first || doesn't even look at the right-hand side. If I swap the 0 and 3 in the last || around, the result remains 3.
The closest I've come is
0 || false && "" || NaN
...which is NaN, whereas
0 || false && ("" || NaN)
...is false, but I think that's explained purely by the left-to-right semantics, not by && being higher precedence.
I must just be missing it, for what expression does it matter that && has a higher precedence than ||?
If they had the same precedence and were left-associative, then e.g. the expression
1 || 0 && 2
would be
((1 || 0) && 2) // evaluates to 2
instead of the
(1 || (0 && 2)) // evaluates to 1
that we get from the "usual" precedence rules.
For your structure … || … && … || … (which would be (((… || …) && …) || …) instead of normal ((… || (… && …)) || …)), you'd get different results for values like 0 0 1 0.
Why does the logical AND have slightly higher precedence to the logical OR?
So that the canonical form of boolean expressions, the disjunctive normal form, does not need any parenthesis.