关于运算符跟表达式副作用

Question

 var a = 1, c;        c = (a++) + a;    console.log(c); // 3问题来了，先执行a++，那a不就等于2了么最后应该是 c = (2) + 2，结果应该是4才对啊，为什么最后会是3++不是有个隐式赋值么，为什么这里的a不会受到副作用影响？

Answer 1

a++是先返回值在进行本身运算的
++a是先进行本身运算在返回值的

所以这个表达式(a++) + a; 运算逻辑为： 先返回(a++)的值为 a等于 1，然后在进行 ++运算这时a=2所以c=3;

如果你的表达式为(++a) + a;那么结果就为4了

Answer 2

c = (a++) + a is equivalent to the following:

var t1 = a;

a++;

var t2 = a;

c = t1 + t2;

In which t1 is 1 and t2 is 2, which is leading to the result 3.

And if you write the code like c = (++a) + a, it will be interpreted as the following:

a++; // or more precisely, ++a

var t1 = a;

var t2 = a;

c = t1 + t2;

And you will get 4;