问题描述
从上一个问题：
Andy Prowl为我提供了这个代码允许我 static_assert 表示模板类型是另一个模板类型：
Andy Prowl provided me with this code that allows me to static_assert that a template type is another template type:
template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of : public std::false_type { };
template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of<TT, TT<Ts...>> : public std::true_type { };
template<typename T>
struct foo {};
template<typename FooType>
struct bar {
static_assert(is_instantiation_of<foo,FooType>::value, ""); //success
};
int main(int,char**)
{
bar<foo<int>> b; //success
return 0;
}
这很好。
但是，如果我改变这样的代码使用 foo 的别名，情况会变坏：
But if I change the code like this to use an alias of foo, things go bad:
template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of : public std::false_type { };
template<template<typename...> class TT, typename... Ts>
struct is_instantiation_of<TT, TT<Ts...>> : public std::true_type { };
template<typename T>
struct foo {};
//Added: alias for foo
template<typename T>
using foo_alt = foo<T>;
template<typename FooType>
struct bar {
//Changed: want to use foo_alt instead of foo here
static_assert(is_instantiation_of<foo_alt,FooType>::value, ""); //fail
};
int main(int,char**) {
//both of these fail:
bar<foo<int>> b;
bar<foo_alt<int>> b2;
return 0;
}
这可以解决吗？
推荐答案
不，它不能被解决（至少不显着改变设计）。问题是，如C ++ 11标准第14.5.7 / 2节所述，不能推导出模板别名：
No, it cannot be solved (not without changing the design significantly at least). The problem is that template alias names are not deduced, as mentioned in paragraph 14.5.7/2 of the C++11 Standard:
示例：
template<class T> struct Alloc { / ... / };
template<class T> using Vec = vector<T, Alloc<T>>;
Vec<int> v; // same as vector<int, Alloc<int>> v;
...
template<template<class> class TT>
void f(TT<int>);
f(v); // error: Vec not deduced <=== Relevant
...
- 结束示例]
在具体情况下，部分专门化，编译器不会推断你的类型是一个实例化 foo_alt （因为 foo_alt 的别名模板），并且选择主模板。
In your concrete case, the problem is that when trying to match the partial specialization, the compiler won't deduce that your type is an instantiation of foo_alt (since foo_alt is the name of an alias template), and the primary template gets picked.
如果要使用别名模板，则必须放弃一些通用性，特定于 foo 的特性：
If you want to use alias templates, you will have to give up a bit of genericity and create a type trait specific for foo:
#include <type_traits>
template<typename T>
struct foo {};
template<typename T>
struct is_instantiation_of_foo : std::false_type { };
template<typename...Ts>
struct is_instantiation_of_foo<foo<Ts...>> : std::true_type { };
那么你可以这样使用：
template<typename FooType>
struct bar {
static_assert(is_instantiation_of_foo<FooType>::value, ""); //fail
};
现在，以下程序中的断言都不会触发：
Now, none of the assertions in the following program will fire:
template<typename T>
using foo_alt = foo<T>;
int main(int,char**) {
// None of these fail:
bar<foo<int>> b;
bar<foo_alt<int>> b2;
return 0;
}
这里是。
这篇关于在模板中使用模板别名，而不是模板的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！