Reputation: 1156
I got a traits class, AttributeTraits, that uses a smart-pointer template class, MySmartPointer :
template<typename T>
class MySmartPointer {};
template<template <typename> class SP = MySmartPointer>
struct AttributeTraits {
template <typename T>
using SmartPointerClass = SP<T>;
};
AttributeTraits use alias to hold the template class it have. Now, I need that Attribute class that gets a trait class will use a typedef/alias to the smart pointer class, such:
template <typename Traits>
struct Attribute {
typedef typename Traits::SmartPointerClass<int> SmartPointerType; // error: non-template 'SmartPointerClass' used as template
using SmartPointerType2 = typename Traits::SmartPointerClass<int>; // error: expected ';' before '<' token
using SmartPointerType3 = Traits::SmartPointerClass<int>; // error: non-template 'SmartPointerClass' used as template
};
But none of the 3 options above works...
Upvotes: 2
Views: 274
Reputation: 41100
This is one of those times where your dependent class is a template, so you have to use the awkward template
syntax when defining your alias:
template <typename Traits>
struct Attribute {
typedef typename Traits::template SmartPointerClass<int> SmartPointerType;
using SmartPointerType2 = typename Traits::template SmartPointerClass<int>;
using SmartPointerType3 = typename Traits::template SmartPointerClass<int>;
};
See this documentation for description of when you need the template
keyword:
The
template
disambiguator for dependent names
Similarly, in a template definition, a dependent name that is not a member of the current instantiation is not considered to be a template name unless the disambiguation keywordtemplate
is used or unless it was already established as a template name
This StackOverflow Answer also provides some good background.
Upvotes: 6