is verifies object identity, and any implementation of Python, when it meets literal of immutable types, is perfectly free to either make a new object of that immutable type, or seek through existing objects of that type to see if some of them could be reused (by adding a new reference to the same underlying … Read more
Not always: >>> nan = float(‘nan’) >>> nan is nan True or formulated the same way as in the question: >>> id(nan) == id(nan) True but >>> nan == nan False NaN is a strange thing. Per definition it is not equal nor less or greater than itself. But it is the same object. More … Read more
You can use is to check if two objects have the same identity. >>> x = [1, 2, 3] >>> y = [1, 2, 3] >>> x == y True >>> x is y False
Another argument against Set being a Functor – it is widely accepted that being a Functor allows you to transform the “elements” of a “collection” while preserving the shape. […] Clearly, any functor instance for Set has the possibility to change the shape, by reducing the number of elements in the set. I’m afraid that … Read more
All types in .NET support the Object.Equals() method which, by default, compares two types for reference equality. However, sometimes, it also desirable to be able to compare two types for structural equality. The best example of this is arrays, which with .NET 4 now implement the IStructuralEquatable interface. This makes it possible to distinguish whether … Read more
You can use Linq: bool areEqual = a.SequenceEqual(b);
In such situation, without rewriting your logic to use named functions, you don’t really have another choice other than declaring the function before the test, e.g. const foo = i => j => i * j const foo2 = foo(2) const bar = () => ({ baz: foo2, boz: 1 }) describe(‘Test anonymous function equality’, … Read more
Try the following var equal = expected.SequenceEqual(actual); Test Version Assert.IsTrue( actual.SequenceEqual(expected) ); The SequenceEqual extension method will compare the elements of the collection in order for equality. See http://msdn.microsoft.com/en-us/library/bb348567(v=vs.100).aspx
if (iar1 == iar2) Here iar1 and iar2 are decaying to pointers to the first elements of the respective arrays. Since they are two distinct arrays, the pointer values are, of course, different and your comparison tests not equal. To do an element-wise comparison, you must either write a loop; or use std::array instead std::array<int, … Read more