As Ryan pointed out, the docs imply that this behavior is intended. If
you want to create a clone that is a result of a merge of several
objects, use Object.merge({}, a, b, c) or resort to the $merge
compatibility function.On Oct 12, 11:46 pm, Perrin Perrin <[email protected]> wrote: > From 1.2 docs: > > var obj1 = {a: 0, b: 1}; > var obj2 = {c: 2, d: 3}; > var obj3 = {a: 4, d: 5}; > var merged = $merge(obj1, obj2, obj3); //returns {a: 4, b: 1, c: 2, d: > 5}, (obj1, obj2, and obj3 are unaltered) > > var nestedObj1 = {a: {b: 1, c: 1}}; > var nestedObj2 = {a: {b: 2}}; > var nested = $merge(nestedObj1, nestedObj2); //returns: {a: {b: 2, c: 1}} > > > > > > > > On Tue, Oct 12, 2010 at 4:44 PM, Ryan Florence <[email protected]> wrote: > > From the 1.3 docs: > > > var obj1 = {a: 0, b: 1}; > > var obj2 = {c: 2, d: 3}; > > var obj3 = {a: 4, d: 5}; > > var merged = Object.merge(obj1, obj2, obj3); // returns {a: 4, b: 1, c: 2, > > d: 5}, (obj2, and obj3 are unaltered) > > > merged === obj1; // true, obj1 gets altered and returned as merged object > > > var nestedObj1 = {a: {b: 1, c: 1}}; > > var nestedObj2 = {a: {b: 2}}; > > var nested = Object.merge(nestedObj1, nestedObj2); // returns: {a: {b: 2, > > c: 1}} > > > On Oct 12, 2010, at 3:39 PM, Oskar Krawczyk wrote: > > > Ouch. > > > Yes, this should not happen. > > > Can you post a ticket on lighthouse? > > > On 2010-10-12, at 23:32, atom wrote: > > > bit of unexpected behavior (which might be intentional). When using > > > Object.merge the objects I pass in are being altered, unlike how > > > $merge used to behave. > > > old way: > > > var obj1 = {a: 1, b: 2, c: 3} > > > var obj2 = {d: 4, e: 5, f: 6} > > > var merged = $merge(obj1, obj2); > > > console.log(obj1); // returns {a: 1, b: 2, c: 3} > > > new way: > > > var obj1 = {a: 1, b: 2, c: 3} > > > var obj2 = {d: 4, e: 5, f: 6} > > > var merged = Object.merge(obj1, obj2); > > > console.log(obj1); // returns {a: 1, b: 2, c: 3, d: 4, e: 5, f: 6}
