equality

I have a class MyCloth and one one object instance of that class that I instantiated like this: MyCloth** cloth1; And at one point in the program, I will do something like this: MyCloth** cloth2 = cloth1; And then at some point later, I want to check to see if cloth1 and cloth2 are the same. (Something like object equality in Java, only here, MyCloth is a very complex class and I can''t build an isEqual function.) How can I do this equality check? I was thinking maybe checking if they point to the same addresses. Is that a good idea? If so, how do I do that? You can test for object identity by

Ok, something strange is happening when writing your own equals operator for NSObject subclasses in Swift 2.0 like this: func ==(lhs: MyObject, rhs: MyObject) -> Bool { return lhs.identifier == rhs.identifier } For a class that looks like this: class MyObject: NSObject { let identifier: String init(identifier: String) { self.identifier = identifier } } This used to work just fine in Swift 1.2 and below. It still kind of works: let myObject1 = MyObject(identifier: "A") let myObject2 = MyObject(identifier: "A") let result = (myObject1 == myObject2) // result is true So far so good, but what if

According to the docs , Array.include? uses the == comparison on objects. I come from Java where such things are (usually) done with .equals() which is easy to override for a particular object. How can I override == in Ruby to allow me to specify the behaviour of Array.include? for my particular object? Many thanks. In Ruby == is just a method (with some syntax sugar on top allowing you to write foo == bar instead of foo.==(bar) ) and you override == just like you would any other method: class MyClass def ==(other_object) # return true if self is equal to other_object, false otherwise end end

I'm trying to use the new (ES6) Map objects in order to represent a map between properties and a value. I have objects in a form similar to: {key1:value1_1,key2:value2_1},..... {key1:value1_N,key2:value2_N} I want to group them based on both their key1 and key2 value. For example, I want to be able to group the following by x and y : [{x:3,y:5,z:3},{x:3,y:4,z:4},{x:3,y:4,z:7},{x:3,y:1,z:1},{x:3,y:5,z:4}] And obtain a Map containing: {x:3,y:5} ==> {x:3,y:5,z:3},{x:3,y:5,z:4} {x:3,y:4} ==> {x:3,y:4,z:4},{x:3,y:4,z:7} {x:3,y:1} ==> {x:3,y:1,z:1} In Python, I'd use tuples as dictionary keys. ES6

I need to compare two numeric values for equality in Javascript. The values may be NaN as well. I've come up with this code: if (val1 == val2 || isNaN(val1) && isNaN(val2)) ... which is working fine, but it looks bloated to me. I would like to make it more concise. Any ideas? Anant Try using Object.is() , it determines whether two values are the same value. Two values are the same if one of the following holds: both undefined both null both true or both false both strings of the same length with the same characters in the same order both the same object both numbers and both +0 both -0 both

Here is a simple generic type with a unique generic parameter constrained to reference types: class A<T> where T : class { public bool F(T r1, T r2) { return r1 == r2; } } The generated IL by csc.exe is : ldarg.1 box !T ldarg.2 box !T ceq So each parameter is boxed before proceeding with the comparison. But if the constraint indicates that "T" should never be a value type, why is the compiler trying to box r1 and r2 ? It's required to satisfy the verifiability constraints for the generated IL. Note that unverifiable doesn't necessarily mean incorrect . It works just fine without the box