[Solved] confusing notation in C++ (OMNeT++)

Question

Let us assume that msg->dup() returned void * — that is, a pointer to void, which means a pointer whose type the compiler doesn’t track. But you may know, e.g. because of documentation on that function, or because certain preconditions have been met, that msg->dup() will return a pointer to CMessage. Before you can use the return value as such, you need to tell the compiler what the type actually is. You do that by casting the void * to CMessage *, which uses the syntax you see.

Accepted Answer

Let us assume that msg->dup() returned void * — that is, a pointer to void, which means a pointer whose type the compiler doesn’t track. But you may know, e.g. because of documentation on that function, or because certain preconditions have been met, that msg->dup() will return a pointer to CMessage. Before you can use the return value as such, you need to tell the compiler what the type actually is. You do that by casting the void * to CMessage *, which uses the syntax you see.