I am using Angular.js 1.3.x. In previous versions of Angular (including 1.3.0-beta5), the following code would copy the properties from the prototype directly to the new object:

function x() {};
x.prototype.logIt = function() {console.log("it")};
var src = new x(); // x has custom properties on the prototype
var dest = {};
angular.copy(src, dest);
dest.logIt(); // "TypeError" in Angular 1.3.0+ 

However, in Angular.js 1.3.0+, the properties from the prototype are totally lost, despite the fact that the migration guide for 1.2 to 1.3 says:

This changes angular.copy so that it applies the prototype of the original object to the copied object. Previously, angular.copy would copy properties of the original object's prototype chain directly onto the copied object.

How do I keep the properties from the prototype?

I am using Angular.js 1.3.x. In previous versions of Angular (including 1.3.0-beta5), the following code would copy the properties from the prototype directly to the new object:

function x() {};
x.prototype.logIt = function() {console.log("it")};
var src = new x(); // x has custom properties on the prototype
var dest = {};
angular.copy(src, dest);
dest.logIt(); // "TypeError" in Angular 1.3.0+ 

However, in Angular.js 1.3.0+, the properties from the prototype are totally lost, despite the fact that the migration guide for 1.2 to 1.3 says:

This changes angular.copy so that it applies the prototype of the original object to the copied object. Previously, angular.copy would copy properties of the original object's prototype chain directly onto the copied object.

How do I keep the properties from the prototype?

• javascript
• angularjs

1 Answer 1

While the ment in the mit linked to from the migration guide says:

This changes angular.copy so that it applies the prototype of the original object to the copied object.

this is only true when the destination argument to angular.copy(source, [destination]); is not provided. When destination is provided, only the direct properties of the object are copied.

The solution is to provide only the source object to the angular.copy function, leaving off the destination parameter:

function x() {};
x.prototype.logIt = function() {console.log("it")};
var src = new x();
var dest = angular.copy(src); // no second parameter
dest.logIt(); // logs "it"

Update: This appears to still be relevant, as angular.copy in v1.6.5 only calls getPrototypeOf(source) if destination is undefined.