图形，顶点，STL列表：为什么列表出来为空？

Question

I posted this question earlier but badly worded which is why it got bad reactions. I will try to be clear in what I am asking. Please comment if you do not understand my question. I am using two lists, one to hold vertices, and another to hold the neighbors of each vertex. When I create a vertex, I push it into the first list. When I want to create edges, I find the vertex in my 'keys' list and add the neighboring vertex into its 'neighbor' list. I repeat for all edges. However, my neighbor list always comes out empty. I check if it did push the vertex into the neighbor list with a print statement in the addEdge function, which shows that it does go in. The neighbor list always come up empty when I try to access them later. Why does that happen? Why does the second list always come empty? Is the memory not preserved?

Eg: Create vertex 1, 2, 3 [stored in keys list]

Create edge: 1-1, 1-2 (2-1), 1-3 (3-1 [stored in neighbor list]

Find edges of 1: nothing is found (why?)

My implementation. The graph class holds the keys list, the vertex holds the neighbor list. The functions that I use to add edges are named: addEdge in both classes. The main also prints that the vertices have no neighbors.

Answer 1

The central issue is that you are passing around values, modifying copies of them and expecting the original to be modified as well. In particular, the issues are

1. You want getV to return a non-const reference to a vertex<T> . Note that the name get is usually associated with a const method that doesn't affect the object. So, you might want to change the name a bit.
2. In graph::addEdge , sKey and dKey should really be vertex<T>& . Also, when you call addEdge on sKey and dKey , pass in the entire vertex<T> object that you want to add, instead of the object of type T . The code, as it stands, compiles because T can be implicitly converted to vertex<T> but that isn't what you want.
3. Lastly, why are you maintaining a list of vertex<T> s in vertex ? Change it to a list<vertex<T>*> s. Note that if you end up changing the data structure in graph from list to something else that could invalidate iterators and pointers to its elements, this approach will fail.