public void DisconnectNodeShouldRemoveEdge(ControlFlowEdgeType edgeType, bool removeSourceNode)
{
var graph = new ControlFlowGraph <int>(IntArchitecture.Instance);
var n1 = new ControlFlowNode <int>(0, 0);
var n2 = new ControlFlowNode <int>(1, 1);
graph.Nodes.AddRange(new[]
{
n1,
n2
});
n1.ConnectWith(n2, edgeType);
if (removeSourceNode)
{
n1.Disconnect();
}
else
{
n2.Disconnect();
}
Assert.Empty(n1.GetOutgoingEdges());
Assert.Empty(n2.GetIncomingEdges());
}
private static ControlFlowNode <TInstruction> GetFallThroughPredecessor <TInstruction>(
ControlFlowNode <TInstruction> node)
{
// There can only be one incoming fallthrough edge for every node. If more than one exists,
// the input control flow graph is constructed incorrectly.
//
// Proof: Suppose there exist two distinct basic blocks v, w that share fallthrough neighbour f,
// then the footers of both v and w must be able to transfer control to the first instruction
// of f without the means of a jump instruction. This can only happen when both footer
// instructions of v and w are placed right before the header of f. Therefore, the footers of
// v and w must be the same instruction, implying v = w, which is a contradiction.
ControlFlowNode <TInstruction> predecessor = null;
foreach (var incomingEdge in node.GetIncomingEdges())
{
if (incomingEdge.Type == ControlFlowEdgeType.FallThrough)
{
if (predecessor != null)
{
throw new BlockOrderingException(
$"Node {node.Offset:X8} has multiple fallthrough predecessors.");
}
predecessor = incomingEdge.Origin;
}
}
return(predecessor);
}
public void RemoveNodeShouldRemoveOutgoingEdges()
{
var graph = new ControlFlowGraph <int>(IntArchitecture.Instance);
var n1 = new ControlFlowNode <int>(1);
var n2 = new ControlFlowNode <int>(2);
graph.Nodes.Add(n1);
graph.Nodes.Add(n2);
n1.ConnectWith(n2);
Assert.Single(n1.GetOutgoingEdges());
Assert.Single(n2.GetIncomingEdges());
graph.Nodes.Remove(n1);
Assert.Empty(n1.GetOutgoingEdges());
Assert.Empty(n2.GetIncomingEdges());
}
