我遇到了一个非常奇怪的编译器错误。由于某种原因,发布的代码确实可以使用 g++ (7.3.0) 正确编译,而 clang (7.0.0) 失败:
../TemplateAlias/main.cpp:64:9: error: no matching function for call to 'freeFunc'
freeFunc(new Func, dummyField);
^~~~~~~~
../TemplateAlias/main.cpp:73:12: note: in instantiation of member function 'Helper<Traits<double, ConcreteData, ConcreteField> >::func' requested here
helper.func();
^
../TemplateAlias/main.cpp:21:13: note: candidate template ignored: deduced conflicting templates for parameter '' ('FieldData' vs. 'ConcreteData')
static void freeFunc(SomeFunc<T, FieldData>* func,
^
两个编译器选项都设置为 -std=c++14
template<typename T>
struct ConcreteData
{
T data;
};
template<typename T, template<typename U> class FieldData>
struct ConcreteField
{
FieldData<T> someMember;
};
template<typename T, template<typename U> class FieldData>
struct SomeFunc
{
};
template<typename T, template<typename U> class FieldData>
static void freeFunc(SomeFunc<T, FieldData>* func,
ConcreteField<T, FieldData>& field)
{
// apply the func on data
(void)field; // silence compiler warning
delete func;
}
template<
typename ScalarType,
template<typename U> class FieldDataType,
template<typename U, template <typename X> class Data> class FieldType
>
struct Traits
{
using Scalar = ScalarType;
template<typename T>
using FieldData = FieldDataType<T>;
using Field = FieldType<Scalar, FieldDataType>; // fails with clang only
// using Field = FieldType<Scalar, FieldData>; // using this line helps clang
};
template<typename Traits>
struct Helper
{
// alias all types given by trait for easier access
using Scalar = typename Traits::Scalar;
using Field = typename Traits::Field;
template<typename U>
using DataAlias = typename Traits::template FieldData<U>;
void func()
{
// using Func = SomeFunc<Scalar, DataAlias>; // this line is intended, but fails with both GCC and clang
using Func = SomeFunc<Scalar, Traits::template FieldData>; // compiles only with GCC, fails with clang
Field dummyField;
freeFunc(new Func, dummyField);
}
};
int main()
{
using ConcreteTraits = Traits<double, ConcreteData, ConcreteField>;
Helper<ConcreteTraits> helper;
helper.func();
return 0;
}
根据 cppreference.com:
类型别名声明引入了一个名称,该名称可以用作由 type-id 表示的类型的同义词。它不会引入新类型,也无法更改现有类型名称的含义。类型别名声明和 typedef 声明之间没有区别。此声明可能出现在块作用域、类作用域或命名空间作用域中。
和
在推导模板模板参数时,别名模板永远不会通过模板参数推导来推导。
据我了解,这两种类型(ConcreteData 和 FieldData)应该是等价的。为什么在这种情况下clang会失败,为什么在使用“第二阶段”别名时两个编译器都会失败?根据 C++ 标准,哪个编译器是正确的?是编译器错误还是对 C++14 标准的微妙含糊解释?