What happens to argument value of first .next() in generator function*

August 25, 2022

Consider this generator function.

Why is the argument of the first call to .next() essentially lost? It will yield and log each of the letter strings but skips "A". Can someone offer an explanation. Please advise on a way that I can access each argument each argument of the .next() method and store it in an array within the generator function?

function* gen(arg) {
  let argumentsPassedIn = [];
  while (true) {
    console.log(argumentsPassedIn);
    arg = yield arg;
    argumentsPassedIn.push(arg);
  }

}


const g = gen();
g.next("A"); // ??
g.next("B");
g.next("C");
g.next("D");
g.next("E");

>Solution :

As per the docs

The first call of next executes from the start of the function until the first yield statement

So when you call the first next, it just calls the generator function from start, where arg will be empty and from next call it works normal.

To make you code work, you should try like this.

function* gen(arg) {
  let argumentsPassedIn = [];
  while (true) {
    console.log(argumentsPassedIn);
    arg = yield arg;
    argumentsPassedIn.push(arg);
  }

}


const g = gen();
g.next()
g.next("A"); // ??
g.next("B");
g.next("C");
g.next("D");
g.next("E");