From: Matthias Kaeppler (nospam_at_[hidden])
Date: 2005-03-04 14:46:24
Ian McCulloch wrote:
>>template< typename Operation >
>>class indirecter_unary
>> : public std::unary_function< typename
>>boost::unary_traits<Operation>::argument_type,
>
>
> Is this correct? You are advertising that your function has an
> argument_type that is the same as Operation::argument_type. Don't you want
> your argument_type to be a pointer?
Hm yes, that is true. So let me correct this to:
class indirecter_unary
: public std::unary_function< typename
boost::unary_traits<Operation>::argument_type*,
typename boost::unary_traits<Operation>::result_type >
> In any event, since you need to
> redefine argument_type and result_type yourself, there is no point
> inheriting from std::unary_function.
>
Why? I thought those types are boost::unary_traits specific types? I
thought I have to inherit from std::unary_function if I want to create
what is called an "Adaptable Unary Function"?
>
>>typename boost::unary_traits<Operation>::result_type >
>>{
>> typedef typename boost::unary_traits<Operation>::argument_type
>> arg_type;
>
>
> I think, the argument_type you are looking for is
>
> typedef typename boost::remove_reference<arg_type>::type* argument_type;
Thanks, that sounds reasonable.
>
>> typedef typename boost::unary_traits<Operation>::param_type
>>param_type;
>
>
> IIUC, param_type is supposed to be a type used to pass this function object
> as a parameter. ie, it should be something like indirecter_unary const&,
> NOT the param_type of Operation.
Yes, I know. That's how I used it: To pass objects of type Operation to
the constructor of indirecter_unary.
>
>> typedef typename boost::unary_traits<Operation>::result_type
>>result_type;
>> typedef typename boost::unary_traits<Operation>::function_type
>>function_type;
>> function_type op_;
>>public:
>> explicit indirecter_unary( param_type op ): op_(op) {}
>> result_type operator() (arg_type *arg) const {
>
> ^^^^^^^^^^
> should be:
> result_type operator()(argument_type arg) const {
>
>
Should it? operator() is supposed to take pointers! Where would the
indirection be if I would pass non-pointer types?
-- Matthias Kaeppler