CODEWITHSUNDEEP
javascriptdart
jcuenod
jcuenod

Reputation: 58435

Parallel.run syntax

I just watched this introduction from the Dart Summit. During the talk, this code was put up:

=> new List.generate(100, (y) => renderLine(y));

I'm pretty sure I understand that line. The arrow function is new to me but okay - it looks a little coffee-esque. The point though, was changing this function to run in parallel which was done like this:

=> Parallel.run(new List.generate(100, (y) => () => renderLine(y)));

Can someone explain the syntax of (y) => () => renderLine(y)?

Upvotes: 4

Views: 71

Answers (3)

lrn
lrn

Reputation: 71783

The (y) => () => renderLine(y) is a function that returns a function. If you write it without the => shorthand, it is the same as:

(y) {
  return () {
    return renderLine(y);
  };
}

This means that the List.generate calls this function 100 times, with different values for y. Each call returns a function that will call renderLine with a different value.

This generates a list of functions (each taking zero arguments). That list is the argument to Parallel.run.

So, the code is just using a quick in-line way to create a list, but it is equivalent to:

var tempList = [];
for (int y = 0; y < 100; y++) tempList.add(() => renderLine(y));
Parallel.run(tempList);

The Parallel.run function expects a list of functions, and will run the functions in parallel, so you get 100 calls to renderLine executed in parallel, each with a different argument.

Upvotes: 3

Tonio
Tonio

Reputation: 1556

I'm guessing (haven't watched the video) that the Parallel.run() function receives a list (or iterable) of no-arg functions to run in parallel. The List.generate() constructor takes an int (length) and a function (which takes one int as an argument) and generates that many items by calling the function for each int in the range 0 .. n-1. So the code new List.generate(100, (y) => renderLine(y)) generates a List of 100 items, each of which is the result of calling renderLine() on each index (that is, renderLine(0), renderLine(1), ... , renderLine(99)).

The code new List.generate(100, (y) => () => renderLine(y)) returns a list of 100 items, aech of which is a function which takes no argumens, and which when called, calls renderLine() with its index in the list. That is, the list is: () => renderLine(0), () => renderLine(1), () => renderLine(2), ..., () => renderLine(99).

I'm guessing Parallel.run() then runs all those functions in parallel, maybe aggregating the results in a list and returning it? If so, then the code Parallel.run(new List.generate(100, (y) => () => renderLine(y))) does something similar to new List.generate(100, (y) => () => renderLine(y)), except in parallel rather than sequentially.

Upvotes: 0

G&#252;nter Z&#246;chbauer
G&#252;nter Z&#246;chbauer

Reputation: 657781

(y) => () => renderLine() is a function which is passed as second parameter to Parallel.run() and this function takes one argument and returns a function which doesn't take any argument.

I don't know why it is done this way.

(x, y) => x + y; is just a shorthand for (x, y) { return x + y; }

Upvotes: 0

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