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Subject: Re: [boost] [outcome] expected, etc. why are these assignable
From: Robert Ramey (ramey_at_[hidden])
Date: 2017-06-02 16:11:38
On 6/2/17 8:42 AM, Peter Dimov via Boost wrote:
> Robert Ramey wrote:
>
>> And wouldn't making these things immutable, eliminate a large part of
>> the whole controversy?
>
> An immutable result/outcome would have no default constructor (because
> it's default-constructed only in cases you need to put the value into it
> later) which would eliminate the motivation for an empty state as
> something into which to default-construct... but that's only a small
> part of the whole controversy.
Right - so it would only eliminate a small part of the whole controversy.
> Niall's argument for empty state does not rest entirely on default
> construction. He argues that it makes Outcome a four-state variant
> instead of a tri-state one and having the fourth state is useful in
> certain scenarios when the function that returns it has a legitimate
> need to report four states.
>
> That is, he wants to be able to express this:
>
> enum class no_difference
> {
> _
> }
>
> expected<std::path, no_difference, std::error_code, std::exception_ptr>
> void return_first_difference( std::path const & p1, std::path const
> & p2 )
> {
> // ...
> }
>
> (or something along these lines, I may not have gotten the example
> entirely correct, but that's the spirit of it.)
OK - I guess then that "expected" wouldn't be consistent with my
expectations for a function result - which for me is the main (or only)
use case. In my world, I would expect
template<class T, class E>
struct expected {...};
expected<std::path, no_difference>
void return_first_difference( std::path const & p1, std::path const & p2 )
{
// ...
}
then use it like this
std::path p1 = "asdfasdfsdf";
std::path p2 = "34234324";
const expected<std::path, no_difference> result =
return_first_difference(p1,p2);
if(result) // convertable to bool?
// convert result to error_code
const std::error_code e(result, error_catagory);
throw ? std::system_error(e)
const std::path path = result;
I don't see why expected has to have more than two states. One for the
legitimate result and one for anything else. The "anything else"
doesn't have to be dealt with inside of "expected". Basically the E
parameter is an "Escape" clause which is the user's responsibility to
define.
> There are different legitimate ways to draw the line; you could say
> three states are it, for more use expected<>, or you could say, four
> states cover the majority of the legitimate cases, so let's have that.
Hmmm - maybe states and parameters are getting confused. An "Escape"
parameter can contain an arbitrary number of states in one parameter.
Robert Ramey
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