Is there a means to define equality for my objects to solve my problem? No not really. There has been some discussion about this on the mailing list. The result is: Build your own Set/Map abstraction on top of Set/Map, which would convert the objects to a primitive value according to your hashing function. Wait … Read more
One can create an empty array, fill it (otherwise map will skip it) and then map indexes to values: Array(8).fill(0).map((_, i) => i * i);
Summarizing the comments: you can’t use yield inside a regular function, so you can’t use yield with forEach. Here an example of “generatorized” foreach: function * foreach (arr, fn) { var i for (i = 0; i < arr.length; i++) { yield * fn(arr[i]) } } function * gen (number) { yield number + 1 … Read more
If your intention is to some other scope, iterate over it, do some other stuff, then be able to iterate over it again later on in that same scope. Then the only thing you shouldn’t try doing is passing the iterator, instead pass the generator: var generator = function*() { yield 1; yield 2; yield … Read more
Try: var g = f(); // this question is over this value. g.next(); // returns: { value: 1, done: false } g.next(123); // returns: { value: 2, done: false } g.next(456); // returns: { value: [123, 456], done: true }
By using yield, generators can be suspended at any point in the control flow of your function, saving your current state of execution (scope & stack). Without generators, this is more complicated: you need to explicitly keep track of the state branching and (especially) looping control structures need to be represented in a functional way, … Read more
What I don’t understand is how this relates to destructuring. Within removeBreakpoint, you can use url, line, and column directly. The destructuring happens when removeBreakpoint is called with an options object; that object’s matching properties are destructured into individual arguments. Is the idea that you permit the ability to pass an object into this function … Read more
if no such option exists, then maybe there is a nice idiomatic one-liner for doing that ? like, using for…of, or similar ? Indeed, there are several ways to convert a Set to an Array: Using Array.from: Note: safer for TypeScript. const array = Array.from(mySet); Simply spreading the Set out in an array: Note: Spreading … Read more
From the Arrow functions examples, (() => “foobar”)() // returns “foobar” So, the function invocation operator should be outside. (() => { //… })(); Sample: http://www.es6fiddle.net/hsb8s1sj/
Believe it or not… Testing for presence of “=>” in string representation of a function is likely the most reliable way (but not 100%). Obviously we can’t test against either of 2 conditions you mentioned — lack of prototype property and lack of [[Construct]] as that might give false positives with either host objects or … Read more