3 May
2004
3 May
'04
4:14 p.m.
"Brian McNamara" <lorgon@cc.gatech.edu> wrote
On Mon, May 03, 2004 at 09:42:16AM -0400, Arkadiy Vertleyb wrote:
On the other hand, since a lambda expression returns a
default-constructed
functor (does it? -- I am unable to verify it right now) the above doesn't make a lot of sence for Lambda... One can just write:
BOOST_TYPEOF(_1 > 15 && _2 < 20) fun;
I don't think so. I am pretty sure that
_1 < 3.3 and _1 < 4.4
have the same _type_ (but they have different "double" _values_ encoded in the run-time structure when constructed).
Yes, thanks for correcting me. This makes one still want to use the BOOST_TYPEOF_ALLOCATE macro in this case. Arkadiy