As per the operator precedence table for JavaScript, I can see that && has higher precedence than ||.
So, for the following code snippet:
let x, y;
let z = 5 || (x = false) && (y = true);
console.log(z); // 5
console.log(x); // undefined
console.log(y); // undefined
I thought that && should be evaluated first and after short-circuiting for the && part, x would be assigned the value false. And then only, || would be tried to be evaluated.
But, from the output of the console.log, I can clearly see that's not the case here.
Can someone please help me what am I missing here?
Thanks in advance.
What the operator precedence of && of || means is that this:
let z = 5 || (x = false) && (y = true);
gets evaluated as:
let z = 5 || ((x = false) && (y = true));
and not as
let z = (5 || (x = false)) && (y = true);
It's not something that indicates when the values on each side of an operator is evaluated. That's done by the mechanics of each individual operator. For example, for ||, it's specified here:
1. Let lref be the result of evaluating LogicalANDExpression.
2. Let lval be ? GetValue(lref).
3. Let lbool be ToBoolean(lval).
4. If lbool is false, return lval.
5. Let rref be the result of evaluating BitwiseORExpression.
6. Return ? GetValue(rref).
It evaluates GetValue(lref) before GetValue(rref) runs.
In other words, operator precedence indicates which tokens are applied to which operator, and in what order, but not how/when a given operator evaluates the values on its left and right side.