Accessing a field of a class after a lambda expression.

Question

I have a weighted graph with its nodes and edges.

Each node contains a LinkedList called edges that stores the edges of this node. Each edge has an weight and a node (node at the other end).

I already did this:

static void removeEdge(Node n1, Node n2)
{
     n1.edges.Remove(n1.edges.First(a => a.node == n2));
     n2.edges.Remove(n2.edges.First(a => a.node == n1));
}

I am trying to do an updateEdge method, that would take that same lambda expression and then do this:

(a => a.node == n2).weight = otherValue;

but I am getting an error. Isn't this allowed? Or am I doing something wrong? From what I've tested the lambda expression seems ok as far as removing the elements, though I'm new at this so I'm pretty lost tbh.

Answer 1

Aren't you trying to do:

(a => a.node == n2).weight

The "=" will make an assignment, the "==" will make an equal comparison! I think this should solve your problem!

Answer 2

Oooooooh!

I just needed to do this:

n1.edges.First(a => a.node == n2).weight = otherValue;
n2.edges.First(a => a.node == n1).weight = otherValue;

Answer 3

I think you're misunderstanding what a lambda is. When you say something like

...First(x => x.bar == 10)

what happens is we generate something like this:

static bool M(X x) { return x.bar == 10; }
...
...  First(new Func<X, bool>( M ) )

That is, we make a method out of the lambda, and then make a delegate out of the method, and then pass the delegate.

Doing something like

(x=>x.bar==10).foo = whatever

is morally the same as doing something like M.foo = whatever, where M is a method. Methods don't have properties, so this is always illegal. Similarly, lambdas don't have properties either. A lambda is just a convenient syntax for a method.