On Tue, Feb 24, 2015 at 7:59 PM, Jonathan S. Shapiro <[email protected]> wrote: > On Tue, Feb 24, 2015 at 1:40 PM, Matt Oliveri <[email protected]> wrote: >> >> On Tue, Feb 24, 2015 at 4:29 PM, Jonathan S. Shapiro <[email protected]> >> wrote: >> > >> > The downforce example, at least, is a trivial lambda having no escaping >> > closure. That one is okay. The upcast example uses a lambda whose >> > closure >> > escapes. That one is *not* okay. >> >> Wait, you're saying downforce would never heap allocate?? How has the >> closure not escaped when we stick (downforce2to1_1 f) into a shared >> data structure? > > The lambda allocated in the downforce example is allocated once, explicitly, > at the definition of downforce. No *call* to downforce introduces a new > escaping closure.
But the lambda closes over the argument f. Every actual argument would need a new closure. Unless I am failing to apply some standard trick I don't know. _______________________________________________ bitc-dev mailing list [email protected] http://www.coyotos.org/mailman/listinfo/bitc-dev
