Documentation problem

[hkaiser]

From IRC:

[13:23]	tiagofg[m]: I noticed that in HPX documentation, in the 
section "2.5.6 Writing single-node HPX applications", in "channels", 
the example that is there has a mistake. When I do "cout << c.get();" 
will fail, because get() return a future.
[13:25]	tiagofg[m]: to obtain the value 42, I think the solution 
may be "cout << c.get().get();" or "cout << c.get(hpx::launch::sync);"

[msimberg]

Would it actually be more canonical to write the example as the following to make it clear that c.get() returns a future (even though it's silly in this tiny example):

hpx::lcos::local::channel<int> c;
hpx::future<int> f = c.get();
// f.is_ready() == false
c.set(42);
// f.is_ready() == true
cout << f.get();

We can have the hpx::launch::sync as well. I'll grab these from tests (or add an example/test if there are no suitable ones) so that we know that the example compiles...