对JavaScript生成器功能感到困惑

Question

Why the value of the final b is not 24 but 18?
I think when function s2 is called the last time, a is 12 and last is 2 , so b should be equal to 12 * 2 = 24 .

 let a = 1, b = 2; function* foo() { a++; yield; b = b * a; a = (yield b) + 3; } function* bar() { b--; yield; a = (yield 8) + b; b = a * (yield 2); } function step(gen) { let it = gen(); let last; return function () { last = it.next(last).value; }; } let s1 = step(foo); let s2 = step(bar); s2(); //b=1 last=undefined s2(); //last=8 s1(); //a=2 last=undefined s2(); //a=9 last=2 s1(); //b=9 last=9 s1(); //a=12 s2(); //b=24 console.log(a, b); 

Answer 1

In the last line of the bar function:

b = a * (yield 2);

the code already ran the a * before running (yield 2) . So it would seem that a is already evaluated at that point.

If you move the multiplication by a to after (yield 2) , then it seems a is evaluated after (yield 2) is run, thereby ensuring you get the most up to date value of a .

So the last line of the bar function could become:

b = (yield 2) * a;

This can be seen in the example below.

 let a = 1, b = 2; function* foo() { a++; yield; b = b * a; a = (yield b) + 3; } function* bar() { b--; yield; a = (yield 8) + b; b = (yield 2) * a; } function step(gen) { let it = gen(); let last; return function () { last = it.next(last).value; }; } let s1 = step(foo); let s2 = step(bar); s2(); //b=1 last=undefined s2(); //last=8 s1(); //a=2 last=undefined s2(); //a=9 last=2 s1(); //b=9 last=9 s1(); //a=12 s2(); //b=24 console.log(a, b);