C++ 自动推断法';s返回类型
我有以下职能:C++ 自动推断法';s返回类型,c++,c++11,type-inference,C++,C++11,Type Inference,我有以下职能: //method takes 2 or more template parameters template <class A1, class A2, class ...Ax> Value<FooMany> getValue() { //note, FooAll's ctor takes std::string and std::initializer_list<std::size_t> FooAll<Item&
//method takes 2 or more template parameters
template <class A1, class A2, class ...Ax>
Value<FooMany> getValue() {
//note, FooAll's ctor takes std::string and std::initializer_list<std::size_t>
FooAll<Item> hairyStructure("abc", { Foo<A1>::getIndex(), Foo<A2>::getIndex(), Foo<Ax>::getIndex() ... } );
return Value<FooMany>(someData, hairyStructure);
}
//method takes only 1 template parameter
template <class A>
Value<FooSingle> getValue() {
//note, FooOne's ctor takes std::string and std::size_t
FooOne<Item> hairyStructure("abc", Foo<A>::getIndex() );
return Value<FooSingle>(someData, hairyStructure);
}
//方法接受2个或多个模板参数
模板
值getValue(){
//注意,FooAll的ctor接受std::string和std::initializer\u列表
FooAll hairyStructure(“abc”),{Foo
谢谢。#包括
#include <type_traits>
template <class A1, class... Ax>
auto getValue()
-> Value<typename std::conditional<sizeof...(Ax) == 0
, FooSingle, FooMany>::type>
{
typename std::conditional<sizeof...(Ax) == 0
, FooOne<Item>
, FooAll<Item>>::type
hairyStructure("abc", { Foo<A1>::getIndex(), Foo<Ax>::getIndex()... } );
return Value<typename std::conditional<sizeof...(Ax) == 0
, FooSingle, FooMany>::type>(hairyStructure);
}
模板
自动获取值()
->价值观
{
typename std::conditional::type
hairyStructure(“abc”,{Foo类型是一回事,但是您如何在函数中编写return
-语句?我不明白为什么会有显式的'FooAll'。它应该在FooAll和FooOne之间变化,对吗?我真的觉得这比使用两个重载更可读。