Question

我想根据内部参数来专门化模板。我正在使用非严格的评估，这使得事情变得困难。

专业化应该基于最少嵌套的模式匹配。例如：

template<typename T>
struct data1;

template<typename T>
struct fun1 {
  using type = data1<T>;
};

template<typename T>
struct fun2;
template<typename T>
struct fun2<data1<T>> {
  using type = data1<T>;
};

fun2<data1<int>> x1;             // this works as expected, T=int
fun2<data1<fun1<int>>>::type x2; // this works as expected, T=fun1<int>
fun2<fun1<int>>::type x3;        // this should be specialized as fun2<data1<int>>, T=int
fun2<fun2<fun1<int>>>::type x4;  // this should be specialized as fun2<data1<int>>, T=int

我该怎么做？

Answer 1

您可以使用模板模板参数：

template<typename T>
struct data1;

template<typename T>
struct fun1 {
  using type = data1<T>;
};

template<typename T>
struct fun2;

template<class T>
struct fun2<data1<T>>{
  using type = data1<T>;
};

template<template<class> class X, class T>
struct fun2<X<T>>
  : fun2<typename X<T>::type>{};

试验：

#include <type_traits>

static_assert(std::is_same<fun2<data1<int>>::type, data1<int>>::value, "fun2<data1<int>>");
static_assert(std::is_same<fun2<data1<fun1<int>>>::type, data1<fun1<int>>>::value, "fun2<data1<fun1<int>>>");
static_assert(std::is_same<fun2<fun1<int>>::type, data1<int>>::value, "fun2<fun1<int>>");
static_assert(std::is_same<fun2<fun2<fun1<int>>>::type, data1<int>>::value, "fun2<fun2<fun1<int>>>");

int main(){
}

Live example on Ideone (with using-aliases changed to typedefs)

基于嵌套内部参数的专用模板

1 个答案: