Joel de Guzman wrote:
Steven Watanabe wrote:
AMDG
Eric Niebler wrote:
The other thorny issue is that lambda expressions aren't Assignable, because their operator= creates a new lambda exression instead. This might well be unfixable.
We could make lambdas Assignable by defining the ordinary assignment operator when the rhs has the same type as the lhs. That would rule out strange lambdas like "_1 = _1" , but I don't see that as a huge loss.
what about a nullary lambda:
vector<int> v1, v2;
(ref(v1) = ref(v2))();
Yikes! Yeah :(
It actually doesn't have to be a problem. Imagine: template<class T> struct actor { // non-const, does assignment actor & operator=(actor const &); // non-const assignment // const, builds an expression template actor</implementation-defined/> const operator=(actor const &) const; }; And then all the operators that build lambda expressions (e.g. phoenix::ref) return const rvalues. So an expression like: (ref(v1) = ref(v2))() would build an expression template, without having to sacrifice Assignability. -- Eric Niebler BoostPro Computing http://www.boostpro.com