Re: Template argument deduction

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Tue, 24 Jan 2012 00:36:13 -0800 (PST)
On 2012-01-24 02:41, kelvSYC wrote:

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<class Value, class Key>
    shared_ptr<Value> operator()(const Key& key) { return
shared_ptr<Value>(new Value(key)); }

OK, so we have here again the same situation as in your thread "The
first template argument" where I explained that this operator() overload
is not very useful and suggested a different design.

Now, consider a class which wraps a std::map<Key, shared_ptr<Value>>
and some kind of delegate class like HeapConvertInserter. One method
which uses the two goes as follows:

// table is the std::map<Key, shared_ptr<Value>> instance, delegate is
the HeapConvertInserter instance
if (table.count(key) == 0) {
    table.insert(std::map<Key, shared_ptr<Value>>::value_type(key,

When I try to compile, apparently it can't deduce the Value template
argument of the HeapConvertInserter's operator(),

Correct, for reasons as explained elsewhere.

even though it would appear that Value should appear as, well, Value
as defined in the std::map instance.

I don't understand how you come to the conclusion that the compiler
understands your intentions?

Why is this happening?

The reasons are the same as before.

Is there a way to make
this work without moving the Value off of the operator() template and
into the class template?

As suggested in the other thread, you could replace the operator()
overload by a member function template (e.g. "call") and invoke it by
providing the return type argument explicitly.

HTH & Greetings from Bremen,

Daniel Kr?gler

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