Incidentally, my preferred way to "return a closure" is to use
an impl like so:
struct ntimes<T>(times: uint, value: T) -> T;
impl<T:Clone> ntimes<T> {
fn call(&mut self) -> Option<T> {
if self.times == 0 {
None
} else {
self.times -= 1;
Some(self.value.clone());
}
}
}
Now you can "call" it like:
let foo = ntimes(3, v);
foo.call(); // Some(v)
foo.call(); // Some(v)
foo.call(); // Some(v)
foo.call(); // None
Niko
On Mon, Oct 28, 2013 at 04:06:38PM -0400, Steven Blenkinsop wrote:
> On Monday, October 28, 2013, Ramakrishnan Muthukrishnan wrote:
>
> > On Mon, Oct 28, 2013 at 11:48 AM, Jesse Ruderman
> > <[email protected]<javascript:;>>
> > wrote:
> > > If you don't mind changing ntimes to not return a closure:
> > >
> > > fn ntimes<T>(f: &fn(T) -> T, times: uint, x: T) -> T {
> > > match times {
> > > 0u => x,
> > > _ => ntimes(|x| f(x), times - 1u, f(x))
> > > }
> > > }
> > >
> > > fn main() {
> > > println(format!("{:d}", ntimes(|k| k*2, 2u, 3)));
> > > }
> >
> > Thanks. That is very nice.
> >
> > No, I am not particular about returning a closure. I was just trying
> > out some toy programs to learn Rust.
> >
> > It would have been nice to return a closure from `ntimes' though. :(
>
>
> You can *create* a closure from ntimes, doing `|x| ntimes(|k| k*2, 2, x)`.
> The idea is that closures borrow data from the stack, so passing them up
> the stack is difficult. If you want to use the stack as your datastructure,
> you have to ensure everything travels down the stack, which you can do by
> requiring the caller to pass in their own closure accepting your closure as
> an argument (a syntax extension abstracting over this pattern would be nice
> so the code doesn't become ungainly). Otherwise, you have to build up your
> own datastructure to pass up the stack, and then derive a closure from that
> somehow. Ultimately, if closures become a special case of trait objects,
> this should be really straightforward to do.
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