[mercury-users] Constructor question.

[Ian MacLarty]

On Sat, Apr 14, 2012 at 9:59 PM, Ian MacLarty
<maclarty at csse.unimelb.edu.au> wrote:
> On Sat, Apr 14, 2012 at 9:42 PM, Michael Richter <ttmrichter at gmail.com> wrote:
>> On 14 April 2012 19:32, Michael Richter <ttmrichter at gmail.com> wrote:
>>>
>>> On 14 April 2012 19:21, Ian MacLarty <maclarty at csse.unimelb.edu.au> wrote:
>>>>
>>>> > You can also do something like this:
>>>>
>>>>
>>>> Foo ^ bar = 10,
>>>> Foo ^ baz = "eleven",
>>>> Foo ^ quux = yes,
>>>
>>>
>>> OK, I … am duly chastened.  I hadn't even thought of doing it that way.  I
>>> was constantly messing with :=.
>>
>>
>> OK, I'm still lost, I'm afraid.  Given this type declaration:
>>
>> :- type foo(bar::int, baz::string, quux::bool).
>>
>> How would I instantiate this using the system you show?
>>
>> :- func default_instance = (foo::out) is det.
>> default_instance = <what code do I put here?>
>>
>> I've tried a few ways to do this, with = or :=, explicitly naming the return
>> variable and separating it from the instantiation code with :- (and a few
>> other more desperate attempts) and I'm just not having any luck.
>>
>
> This works for me:
>
> :- type foo --->  foo(bar :: int, baz :: string).
> :- func default_instance = (foo::out) is det.
> default_instance = Foo :-
>    Foo ^ bar = 10,
>    Foo ^ baz = "eleven".
>

As does this:

:- type foo --->  foo(bar :: int, baz :: string, quux :: bool).
:- func default_instance = (foo::out) is det.
default_instance = Foo :-
    Foo ^ bar = 10,
    Foo ^ baz = "eleven",
    Foo ^ quux = yes.

If it doesn't work for you what error do you get?

Ian.

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