最后编辑
我有一个带有模板的函数:
template <template <typename ...> class P, typename ... Args>
void f(const P<Args...> &p)
{
std::cout << "Template with " << sizeof...(Args) << " parameters!\n";
}
它适用于我迄今为止测试过的任何类型的模板:
f(std::valarray<int>{}); // Prints: "Template with 1 parameters!"
f(std::pair<char, char>{}); // Prints: "Template with 2 parameters!"
f(std::set<float>{}); // Prints: "Template with 3 parameters!"
f(std::map<int, int>{}); // Prints: "Template with 4 parameters!"
但是,假设我想专门化模板,当它需要一个带有两个参数的模板时,下面的代码不起作用:
template <>
void f<template <typename, typename> class P, typename A, typename B>(const P<A, B> &p)
{
std::cout << "Special case!\n";
}
parse error in template argument list
void f<template <typename, typename> class P, typename A, typename B>(const P<Args...> &p) { std::cout << "Template with " << sizeof...(Args) << " parameters!\n"; }
^
'P' does not name a type
void f<template <typename, typename> class P, typename A, typename B>(const P<Args...> &p) { std::cout << "Template with " << sizeof...(Args) << " parameters!\n"; }
^
expected ',' or '...' before '<' token
void f<template <typename, typename> class P, typename A, typename B>(const P<Args...> &p) { std::cout << "Template with " << sizeof...(Args) << " parameters!\n"; }
^
template-id 'f<<expression error> >' for 'void f(int)' does not match any template declaration
void f<template <typename, typename> class P, typename A, typename B>(const P<Args...> &p) { std::cout << "Template with " << sizeof...(Args) << " parameters!\n"; }
AFAIK 使用其他类型的模板参数非常简单:
// Type parameter
template <typename TYPE>
void f(TYPE) { std::cout << "Type!\n"; }
// Non-type parameter
template <int VALUE>
void f() { std::cout << "Value " << VALUE << "!\n"; }
// Type parameter specialization
template <>
void f(float) { std::cout << "Special type case!\n"; }
// Non-type parameter specialization
template <>
void f<666>() { static_assert(false, "Forbidden number!"); }
如何使用模板模板模板实现此功能?
编辑:
正如orlp和angew函数模板所指出的那样,不能部分专门化,所以我应该坚持使用对象模板,这是我的尝试:
template <template <typename ...> class P, typename ... Args>
struct c
{
using type = P<Args...>;
const std::size_t count = sizeof...(Args);
void f(const type &t)
{
std::cout << "Template with " << sizeof...(Args) << " parameters!\n";
}
};
template <template <typename, typename> class P, typename A, typename B>
struct c<P, A, B>
{
using type = P<A, B>;
void f(const type &t)
{
std::cout << "Spezialized 2 parameters!\n";
}
};
有用:
c<std::valarray, int> c_valarray_int;
c<std::pair, int, char> c_pair_int_char;
c<std::vector, int, std::allocator<int>> c_vector_int;
c<std::map, int, int> c_map_int_int;
c_valarray_int.f({}); // Prints: "Template with 1 parameters!"
c_pair_int_char.f({}); // Prints: "Spezialized with 2 parameters!"
c_vector_int.f({}); // Prints: "Spezialized with 2 parameters!"
c_map_int_int.f({}); // Prints: "Template with 2 parameters!" (expected)
但是现在,我应该指定所有参数,而不是让编译器猜测整个事情,嗯……这不是悲剧。