Why is the Javascript operator "&&" so weird?

Question

a = 1;
b = "1";
if (a == b && a = 1) {
    console.log("a==b");
}

The Javascript code above will result in an error in the if statement in Google Chrome 26.0.1410.43:

Uncaught ReferenceError: Invalid left-hand side in assignment

I think this is because the variable a in the second part of the statement &&, a=1 cannot be assigned. However, when I try the code below, I'm totally confused!

a = 1;
b = "1";
if (a = 1 && a == b) {
    console.log("a==b");
}

Why is the one statement right but the other statement wrong?

Answer 1

= has lower operator precendence than both && and ==, which means that your first assignment turns into

if ((a == b && a) = 1) {

Since you can't assign to an expression in this way, this will give you an error.

Answer 2

The second version is parsed as a = (1 && a == b); that is, the result of the expression 1 && a == b is assigned to a.

The first version does not work because the lefthand side of the assignment is not parsed as you expected. It parses the expression as if you're trying to assign a value to everything on the righthand side--(a == b && a) = 1.

This is all based on the precedence of the various operators. The problem here stems from the fact that = has a lower precedence than the other operators.