JavaScript运算顺序疑问：括号为何未优先于后自增运算符执行？

Great question—this is a super common confusion when working with JavaScript's increment operators, so let's break it down clearly.

First, let's confirm both examples give the exact same result:

• Example 1:

  var a = 0; var b = 1; a = (b++); console.log(a); // Logs 1
  

• Example 2:

  var a = 0; var b = 1; a = b++; console.log(a); // Also logs 1
  

The Key Misconception: Parentheses vs. Operator Behavior #

You’re right that parentheses have the highest operator priority, but here’s the critical detail: parentheses only change how expressions are grouped—they don’t alter the fundamental behavior of the operator inside them.

The post-increment operator (x++) has fixed, unchanging rules, no matter if you wrap it in parentheses or not:

1. First, return the current value of x as the result of the expression.
2. Then, increment x by 1 (this happens after the value is returned).

Let’s Walk Through Example 1 Step-by-Step #

When JavaScript executes a = (b++):

1. The parentheses tell JavaScript to evaluate b++ as a single, standalone expression first (which it would do anyway here, since ++ has higher priority than =).
2. Evaluating b++ returns the current value of b (which is 1) and then increments b to 2.
3. That returned value (1) is assigned to a.

What If We Used Pre-Increment? #

If you had used the pre-increment operator (++x) instead, parentheses would still follow the operator’s behavior:

var a = 0; var b = 1; a = (++b); console.log(a); // Logs 2

Here, ++b first increments b to 2, then returns that new value—so a gets 2. Parentheses don’t change this, they just group the expression.

To Sum It Up #

• Post-increment (x++): Always returns the current value first, then increments. Parentheses don’t override this.
• Pre-increment (++x): Always increments first, then returns the new value.
• Parentheses control grouping and evaluation order, but never change an operator’s built-in behavior.

内容的提问来源于stack exchange，提问作者Lisa Friedrichsen