C# – What Do Braces After New Statement Do?

.netc++new-operator

Given the code below, what is the difference between the way position0 is initialized and the way position1 is initialized? Are they equivalent? If not, what is the difference?

class Program
{
    static void Main(string[] args)
    {
        Position position0 = new Position() { x=3, y=4 };

        Position position1 = new Position();
        position1.x = 3;
        position1.y = 4;
    }
}

struct Position
{
    public int x, y;
}

Best Answer

They are not quite equivalent - at least not in the general case. The code using an object initializer is closer to this:

Position tmp = new Position();
tmp.x = 3;
tmp.y = 4;
Position position1 = tmp;

In other words, the assignment to the variable only occurs after the properties have been set. Now in the case where you're declaring a new local variable, that doesn't actually matter, and the compiler may well optimize to your first form. But logically, it does matter. Consider:

Position p1 = new Position { x = 10, y = 20 };

p1 = new Position { x = p1.y, y = p1.x };

If that did the assignment to p1 first, you'd end up with 0 for both p1.x and p1.y. Whereas that's actually equivalent to:

Position tmp = new Position();
tmp.x = 10;
tmp.y = 20;
Position p1 = tmp;

tmp = new Position();
tmp.x = p1.y; // 20
tmp.y = p1.x; // 10
p1 = tmp;

EDIT: I've just realised that you're using a struct rather than a class. That may make some subtle differences... but you almost certainly shouldn't be using a mutable struct to start with :)

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