This answer is based off this fun-fact: In a function returning void
, you can return any expression of which the type is void.
So the simple solution is:
virtual T Execute() const
{
if (comm) // boolean logic change, typo in OP?
return comm();
else
return static_cast<T>(NULL);
}
When T = void
, the last return statement is equivalent to return;
.
However, I feel this is bad design. Is NULL
meaningful for every T
? I don't think so. I would throw an exception:
virtual T Execute() const
{
if (comm)
return comm();
else
throw std::runtime_error("No function!")
}
However, this is done automatically by Boost, so your code becomes the much cleaner:
virtual T Execute() const
{
return comm();
}
You could then add additional functionality, such as:
bool empty(void) const
{
return !comm; // or return comm.empty() if you're the explicit type
}
So the user can check if it can be called prior to calling it. Of course at this point, unless your class has additional functionality you've left out for the sake of the question, I see no reason not to just use boost::function
in the first place.