C++17 指向带有类模板默认参数的模板类的指针

Question

I am experimenting with C++17 class template default argument and was wondering if anyone could explain:

If I have the following:

template<typename Policy = DefaultPolicy>
class MyClass { //class goes here };

And then try to use it as:

MyClass * class = new MyClass();

I get the error:

However both the following compile OK:

MyClass stackClass = MyClass();

auto * heapClass = new MyClass();

In particular I am very interested in how auto is working above. Thanks so much!

Perhaps there is also a concept name that describes this that I can google for more info also.

Working example: https://godbolt.org/z/EbEnxjcej

Answer 1

Correct syntax forming a pointer to a template instance with default parameter would be:

MyClass<> * heapClass = new MyClass();  
auto smartClass = std::make_unique<MyClass<>>(); // for same reason

MyClass formally isn't a type-id, it's a template name. That's why make_unique would fail, because its parameter should be a typename . Pointer declaration syntax would require same. What auto does is use of a full type-id - MyClass<DefaultPolicy> . The new expression is one of special cases allowed in C++17 along with MyClass stackClass although for clarity new MyClass<>() can be used as pre-17.