Re: [boost] Boost.Local Review (Nov 10, 2011 to Nov 19, 2011)

Subject: Re: [boost] Boost.Local Review (Nov 10, 2011 to Nov 19, 2011)
From: Jeffrey Lee Hellrung, Jr. (jeffrey.hellrung_at_[hidden])
Date: 2011-11-18 18:27:08

On Fri, Nov 18, 2011 at 3:20 PM, Lorenzo Caminiti <lorcaminiti_at_[hidden]>wrote:

> On Fri, Nov 18, 2011 at 1:31 PM, Jeffrey Lee Hellrung, Jr.
> <jeffrey.hellrung_at_[hidden]> wrote:
> > On Fri, Nov 18, 2011 at 1:44 AM, Lorenzo Caminiti <lorcaminiti_at_[hidden]
> >wrote:
> >
> >> On Wed, Nov 16, 2011 at 10:24 AM, Lorenzo Caminiti
> >>
> > [...]
> >
> >> boost::function<void (int&)> counter() {
> >>    int delta = 2;
> >>    void BOOST_CLOSURE(int& x, bind delta) {
> >>
> >
> > Of course, you meant "bind x" or "int x" here instead of "int& x", right?
>
> No, I meant int& x. In this case the local function accepts one
> parameter of type int& and it is closed over a copy of value of the
> local variable delta. Given that none of the variables is bound by
> reference (the only reference is a function parameter and not a bound
> variable), such a closure can be safely returned to another scope and
> executed outside the local scope (within main in this case). When main
> will execute the closure, it will have to pass the one parameter x
> (see inc(x) below).
>
> >>        x += delta;
> >>        std::cout << "x = " << x << std::endl;
> >>    } BOOST_CLOSURE_END(increment)
> >>    return increment;
> >> }
> >>
> >> int main() {
> >>    int x = 1;
> >>    boost::function<void (int&)> inc = counter();
> >>    inc(x); // undefined behaviour now but it should work...
> >>    return 0;
> >> }
> >>
> > [...]
>

Ack, right.  I wasn't paying attention and didn't notice that you had moved
the x variable to main relative to the preceding (snipped) example.

- Jeff