On Friday, 23 November 2018 at 11:29:24 UTC, Nicholas Wilson
wrote:
No, std.functional.memoize uses a hashtable to cache the
runtime results of calls to expensive functions.
assuming that the example is not oversimplified and
generateFunc1 and generateFunc2 are functions, the compiler
doesn't do extra semantic analysis so the validity of the
functions is effectively cached.
If they are templates (with parameters) then the compiler will
automatically memoize them (it too keeps a hashtable of
template instances).
Ah, that's good to know.
