Why is x not equal to one in if (6 || 5 && (x = 1))?

Question

JavaScript

var x = 0
if (6 || 5 && (x = 1)) document.write(x); // 0

As far as I know, the '&&' operator has higher precedence than the '||' operator.
However, the right operand is not executed, why?
PS: Issue resolved. To understand the answer, please read the comments to the solution. Parentheses have nothing to do with it (and operator precedence too).

Answer 1

because, in fact, your expression looks like this:

var x = 0
if (6 || (5 && (x = 1))) document.write(x); // 0

logical operator || works like this: if the first operand can be cast to true, the rest of the expression is not evaluated.
https://tc39.github.io/ecma262/#sec-binary-logical...
the second part of clause 12.13.3 concerns just the case indicated at the beginning of the discussion.

Answer 2

6 evaluates to true, the rest of the expression is simply ignored as having no effect on the result

Answer 3

js interpreter first arranges brackets according to operator precedence, and then evaluates expressions. it turned out that the brackets were placed like this , and then the interpreter realized that the right part || do not need to perform (6 || (5 && (x = 1)))

Answer 4

Maybe because you do not compare x with 1, but assign a unit to x?

Answer 5

These synthetic questions from interviews infuriate ...
Because, for starters, the person who writes such code needs to be beaten on the hands with a ruler.

Answer 6

It's the same as writing if(6) {}
If the first part of the condition returns anything, then the other part is ignored.
To check also the second part of the condition, you need to either compare 6 with something, or instead of || (or) write && (and)

Answer 7

but he just didn’t go further than 6, because he brought it to true