It is important to realise that everything in C# is passed by value, unless you specify ref or out in the signature. What makes value types (and hence structs) different from reference types is that a value type is accessed directly, while a reference type is accessed via its reference. If you pass a reference … Read more
In C++11, you should add rvalue reference: A(A&&); void operator=(A&& a );
When in doubt, pass by value. Now, you should only rarely be in doubt. Often values are expensive to pass and give little benefit. Sometimes you actually want a reference to a possibly mutating value stored elsewhere. Often, in generic code, you don’t know if copying is an expensive operation, so you err on the … Read more
When you call a function or a method and you pass parameters to it, a copy is made from the values, and the function can only access these copies. This means if the function attempts to modify / change the copies, it will not change the original value. For example: func main() { i := … Read more
You can use [:], but for list containing lists(or other mutable objects) you should go for copy.deepcopy(): lis[:] is equivalent to list(lis) or copy.copy(lis), and returns a shallow copy of the list. In [33]: def func(lis): print id(lis) ….: In [34]: lis = [1,2,3] In [35]: id(lis) Out[35]: 158354604 In [36]: func(lis[:]) 158065836 When to … Read more
What I don’t understand is what happens when I invoke a method, what actually happens. Does new() get invoked? Does it just automagically copy the data? Or does it actually just point to the original object? And how does using ref and out affect this? The short answer: The empty constructor will not be called … Read more
In Python a variable is not a box that holds things, it is a name that points to an object. In your code: d = { ‘m’: [1,2,3] } –> binds the name d to a dictionary m = d[‘m’] –> binds the name m to a list m = m[1:] –> binds the name … Read more
A compiler vendor would typically implement a reference as a pointer. Pointers tend to be the same size as or larger than many of the built-in types. For these built-in types the same amount of data would be passed whether you passed by value or by reference. In the function, in order to get the … Read more
As Rytmis said, Java passes references by value. What this means is that you can legitimately call mutating methods on the parameters of a method, but you cannot reassign them and expect the value to propagate. Example: private void goodChangeDog(Dog dog) { dog.setColor(Color.BLACK); // works as expected! } private void badChangeDog(Dog dog) { dog = … Read more
To elaborate the two answers already given a bit: When you define variables to be “the same as” some other variable, you have basically two possibilities: ClassA aCopy = someOtherA; //copy ClassA& aRef = someOtherA; //reference Instead of non-const lvalue references, there are of course const references and rvalue references. The main thing I want … Read more