如何指定模板参数是类模板，并从另一个模板参数推断其模板类型？

问题描述

考虑一个模板函数：

template <typename OutputContainerType, typename ContainerType>
static OutputContainerType processContainer(ContainerType c)
{
    OutputContainerType result;
    ...
    return result;
}

我可以这样说没有问题：

I can call it no problem like so:

std::vector<MyClass> v;
const auto result = processContainer<std::set<MyClass>>(v);

。所以必须指定 std :: set< MyClass>> 是多余的;我想键入 processContainer< std :: set>（v）并且具有推断项类型为 decltype（v）:: value_type 。我该怎么办？我尝试过不同的事情，如

However, I know that the function will accept and produce different containers, but always with the same element type. So having to specify std::set<MyClass>> is redundant; I want to type processContainer<std::set>(v) and have the function infer the item type as decltype(v)::value_type. How can I do that? I've tried different things like

template <template<> class OutputContainerType, class ContainerType>
static OutputContainerType<typename ContainerType::value_type> processContainer(ContainerType c) {}

但不能让它编译无论什么的C ++模板语法和技巧不是很深，如你所见）。

but can't get it to compile no matter what (my understanding of C++ template syntax and tricks is not very deep, as you can see).

推荐答案

如果你不在乎分配器，您可以忽略它：

If you don't care about the allocator, you can just omit it:

template <template<typename...> class OutputContainerType, template<typename...> class ContainerType, typename ValueType>
static OutputContainerType<ValueType> processContainer(ContainerType<ValueType> c)
{
    OutputContainerType<ValueType> result;
    //  ...
    return result;
}

int main() {
    std::set<int> s {1, 2, 3};
    auto v = processContainer<std::vector, std::set, int>(s);
}

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