I'm trying to implement this code:
const a = [1, 2, 3, 4, 5];
a.multiply();
console.log(a); // [1, 2, 3, 4, 5, 1, 4, 9, 16, 25]
The result array needs to be concatenated at the end of the original array. I'm using the following code but it returns only the array with the values multiplied:
Array.prototype.multiply = function() {
return this.map(function(item){return item * item})
}
a.multiply();
console.log(a); //[1, 4, 9, 16, 25]
Can anyone help me please?
You could push the items of mapping to this.
Array.prototype.multiply = function () {
this.push(...this.map(v => v * v)); // push spreaded values
return this; // allow fluent interface
};
var a = [1, 2, 3, 4, 5];
a.multiply();
console.log(a);Pardon me for asking but why mutate the array? You can decorate the Array prototype with functions which simply return the result you need. Mutating the input/state has its disadvantages.
An approach where you do not mutate the array would be something like this:
Array.prototype.multiplyAll = function() {
return this.reduce((r,c) => r * c)
}
Array.prototype.multiplyEach = function() {
return this.map(x => x * x)
}
var a = [1, 2, 3, 4, 5];
console.log(a.multiplyAll());
console.log(a.multiplyEach());In general the recommended approach is not to decorate the prototypes of the build in Objects but to have a utility of pure functions with which you work with.
You can use "Array.concat" and "Array.map" like below
const a = [1, 2, 3, 4, 5];
// for direct method
function multiply() {
return a.concat(a.map(d => d * d))
}
console.log(multiply(a));
// for prototype method
Array.prototype.multiply = function() {
Object.assign(this, this.concat(this.map(d => d * d)))
}
a.multiply()
console.log(a)If you're coding to the ES6 spec you can use spread syntax in conjunction with Array.prototype.map method to achieve this. I re-named the method to pushSquared since I thought it made more sense.
Array.prototype.pushSquared = function() {
Object.assign(this, [...this, ...this.map(n => n * n)]);
}
var a = [1, 2, 3, 4, 5];
a.pushSquared()
console.log(a);
a as asked in OP