26f0200d
From: ulrich.eckhardt@dominolaser.com
Am 24.01.2012 02:41, schrieb kelvSYC:
> I'm having a problem where it would seem like template arguments could
> be deduced, but in reality the compiler barfs because it can't.
> Suppose I have this generic functor:
>
> struct HeapConvertInserter {
> template
> shared_ptr operator()(const Key& key) { ... }
> };
For the record, "Value" will not be deduced here, because it is not used
in the argument list of the function. There are several things that the
compiler could use:
1. The implementation of operator()
This won't be used because in the place where it is used, that
implementation might be unavailable. Also, there is nothing that keeps
two different instantiations from having the same implementation.
2. The assignee of the result
Since you can always ignore returned values, this isn't guaranteed to be
present. Also, if the assignee has a template assignment operator, you
would be lost again.
> Now, consider a class which wraps a std::map>
> and some kind of delegate class like HeapConvertInserter. One method
> which uses the two goes as follows:
>
> /* table is the std::map> instance, delegate is
> the HeapConvertInserter instance */
> if (table.count(key) == 0) {
> table.insert(std::map>::value_type(key,
> delegate(key)));
> }
>
> When I try to compile, apparently it can't deduce the Value template
> argument of the HeapConvertInserter's operator(), even though it would
> appear that Value should appear as, well, Value as defined in the
> std::map instance. Why is this happening?
No, sorry. You can reuse the name "Value" in many different places and
using it in one template won't automatically affect another.
> Is there a way to make this work without moving the Value off of the
> operator() template and into the class template?
Yes, see recent thread you started yourself, i.e. using explicit
template arguments. Note that if you don't want to repeat yourself, you
could reference the type inside the map using
key_type/value_type/element_type typedefs.
Uli
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