If I have a lambda which captures all automatic variables by reference ([&] {}
), why can't it be converted to a function pointer? A regular function can modify variables just like a lambda that captures everything by reference can, so why is it not the same?
I guess in other words, what is the functional difference between a lambda with a &
capture list and a regular function such that the lambda is not convertible to a function pointer?
So let's take the example of a trivial lambda:
What does the type of
foo
look like? It might look something like this:Can you create a function pointer to that
operator()
? No, you can't. That function needs some extra information that comes from its object. This is the same reason that you can't convert a typical class method to a raw function pointer (without the extra first argument wherethis
goes). Supposing you did create some this pointer - how would it know where to geto
from?The "reference" part of the question is not relevant - if your lambda captures anything, then its
operator()
will need to reference some sort of storage in the object. If it needs storage, it can't convert to a raw function pointer.References, though they aren't objects, need to be stored somewhere. A regular function cannot access local variables of another function; Only references (e.g. as parameters) that could refer to local variables. A Lambda with a
&
as the capture-default can, because every variable needed can be captured.In other words: A regular function doesn't have state. A closure object with captured variables does have state. So a closure object cannot be reduced to a regular function, because the state would be lost.