JS: Does Object.assign() create deep copy or shallow copy

Question

I just came across this concept of

var copy = Object.assign({}, originalObject);

which creates a copy of original object into the \"

Answer 1

As mentioned above, Object.assign() will do a shallow clone, fail to copy the source object's custom methods, and fail to copy properties with enumerable: false.

Preserving methods and non-enumerable properties takes more code, but not much more.

This will do a shallow clone of an array or object, copying the source's methods and all properties:

function shallowClone(src) {
  let dest = (src instanceof Array) ? [] : {};

// duplicate prototypes of the source
  Object.setPrototypeOf(dest, Object.getPrototypeOf(src));

  Object.getOwnPropertyNames(src).forEach(name => {
    const descriptor = Object.getOwnPropertyDescriptor(src, name);
    Object.defineProperty(dest, name, descriptor);
  });
  return dest;
}

Example:

class Custom extends Object {
  myCustom() {}
}

const source = new Custom();
source.foo = "this is foo";
Object.defineProperty(source, "nonEnum", {
  value: "do not enumerate",
  enumerable: false
});
Object.defineProperty(source, "nonWrite", {
  value: "do not write",
  writable: false
});
Object.defineProperty(source, "nonConfig", {
  value: "do not config",
  configurable: false
});

let clone = shallowClone(source);

console.log("source.nonEnum:",source.nonEnum);
// source.nonEnum: "do not enumerate"
console.log("clone.nonEnum:", clone.nonEnum);
// clone.nonEnum: – "do not enumerate"

console.log("typeof source.myCustom:", typeof source.myCustom);
// typeof source.myCustom: – "function"
console.log("typeof clone.myCustom:", typeof clone.myCustom);
// typeof clone.myCustom: – "function"

jsfiddle