Object rest spread is stage 3 proposal and not a part of any spec (will probably be included in ES2018). More importantly, export has syntax that mimics existing JS syntax but doesn’t interpret { … } as an expression. export syntax was strictly defined because ES2015 modules are supposed to be statically analyzed. This is … Read more
In your example given, there is essentially no difference between the two .concat is significantly more efficient: http://jsperf.com/spread-into-array-vs-concat because … (spread) is merely syntax sugar on top of more fundamental underlying syntax that explicitly iterates over indexes to expand the array. Spread allows sugared syntax on top of more clunky direct array manipulation To expand … Read more
You can just use inheritance to do that : interface IA { one: number; two: number; } interface IC { other: number; four: number; } interface IB extends IA, IC { three: number; }
Although the other answers are correct, they change the function signature to accept an object instead of 2 separate arguments. Here is how to use an object’s values as function arguments without altering the function’s signature. This requires Object.values (ES 2017) and the spread operator to be available in your runtime. const args = { … Read more
So, for this problem, you have to understand what is the shallow copy and deep copy. Shallow copy is a bit-wise copy of an object which makes a new object by copying the memory address of the original object. That is, it makes a new object by which memory addresses are the same as the … Read more
You can use a combination of spread and Array#slice: const arr = [‘a’, ‘b’, ‘c’, ‘d’, ‘e’]; const indexToRemove = 2; // the ‘c’ const result = […arr.slice(0, indexToRemove), …arr.slice(indexToRemove + 1)]; console.log(result); Another option is Array#filter: const arr = [‘a’, ‘b’, ‘c’, ‘d’, ‘e’]; const indexToRemove = 2; // the ‘c’ const result = … Read more
Both Function.prototype.apply and the spread syntax may cause a stack overflow when applied to large arrays: let xs = new Array(500000), ys = [], zs; xs.fill(“foo”); try { ys.push.apply(ys, xs); } catch (e) { console.log(“apply:”, e.message) } try { ys.push(…xs); } catch (e) { console.log(“spread:”, e.message) } zs = ys.concat(xs); console.log(“concat:”, zs.length) Use Array.prototype.concat instead. … Read more
First of all, Statement A & Statement B will generate different results. In Statement A, you are inserting an array ([“Lemon”, “Kiwi”]) as an array element at position 2 while removing 0 items. So, you are inserting a string array in another string array at position 2. var fruits = [“Banana”, “Orange”, “Apple”, “Mango”]; fruits.splice(2,0,[“Lemon”, … Read more