是否可以分别专注于 typedef 及其基础类型？

Question

简单的例子是size_t。假设：

typedef uint64_t size_t;

而且我想专注于 uint64_t 和 size_t：

template <class T>
struct MyType {};

template <>
struct MyType<uint64_t>{
    void operation()() { cout << "uint64_t" << endl; }
};

template <>
struct MyType<size_t>{
    void operation()() { cout << "size_t" << endl; }
}

MyType<1ULL>()();
MyType<static_cast<size_t>(1ULL)>()();

这可能吗？是否有获得相同净效果的技巧？

Answer 1

不可能。

标准（7.1.3，[dcl.typedef]）内容如下：

A name declared with the typedef specifier becomes a typedef-name. Within the scope of its declaration, a typedef-name is syntactically equivalent to a keyword and names the type associated with the identifier in the way described in Clause 8. A typedef-name is thus a synonym for another type. A typedef-name does not introduce a new type the way a class declaration (9.1) or enum declaration does.

是否可以分别专注于 typedef 及其基础类型？

Is it possible to specialize on a typedef and its underlying type separately?

c++

templates

typedef

template-specialization