std：map作为模板参数的模板推导失败 [英] Template deduction fails for std:map as template parameter

问题描述

我正在C ++ 11中实现一个简单的LRU缓存。我几乎涵盖了所有内容，但是只有一个小问题。假设我有以下模板类定义：

I'm implementing a simple LRU cache in C++11. I pretty much have it covered but there's just one minor problem. Let's say I have the following template class definition:

#ifndef _LRU_STL_H_
#define _LRU_STL_H_

#include <functional>
#include <cassert>
#include <list>

template <typename KeyType, typename ValueType, template<typename...> class Map>
class LRU {

public:
    typedef Map<KeyType, std::pair<ValueType, typename std::list<KeyType>::iterator>> KeyToValueType;

    LRU(const std::function<ValueType(const KeyType&)> &Function, size_t Capacity)
    : _Function(Function), _Capacity(Capacity) {
        assert(_Capacity != 0);
    }

    ...

private:
    ...

    std::function<ValueType(const KeyType&)> _Function;
    const size_t _Capacity;
    KeyToValueType _KeyToValue;
};

#endif

在KeyToValue类型下，MSVC2013出现以下编译错误：错误1错误C2976： std :: map：模板参数过少c：\x\visual studio 2013\项目\缓存\lru_stl\lru_stl.h 17 1 LRU_STL

At KeyToValue type I get the following compilation error with MSVC2013: Error 1 error C2976: 'std::map' : too few template arguments c:\x\visual studio 2013\projects\caching\lru_stl\lru_stl.h 17 1 LRU_STL

第17行是：

typedef Map<KeyType, std::pair<ValueType, typename std::list<KeyType>::iterator>> KeyToValueType;

似乎模板推导失败。这可能是一个非常简单的问题，但我仍然找不到。为了完整起见，下面是一个示例：

Seems like the template deduction fails. It may be a very simple problem but I just couldn't find it yet. For completeness here's an example:

std::function<std::string(std::string)> functionToCache = [](std::string & str) {
    std::string reverse;
    reverse.reserve(str.size());

    std::copy(str.begin(), str.end(), reverse);
    return reverse;
};

LRU<std::string, std::string, std::map> LRU(functionToCache, 5);
std::string Hello_World = LRU("Hello World");
assert(Hello_World == "dlroW olleH");

已提供错误。做过修复。仍然发生相同的错误：std :: map模板参数太少。

The error is already provided. Done mentioned fixes. Still the same error occurs: std::map too few template arguments.

为了完整性，如果我删除所有内容并创建一个TEST类：

Just for completeness if I remove everything and create a TEST class:

template <typename A, typename B, template <typename ...> class Map>
class TEST {
    typename Map<A, std::pair<B, typename std::list<A>::iterator>> CMAP;
public:
    TEST(void) { }
};

尝试实例化该类会导致完全相同的错误消息。

Trying to instantiate the class results in the exact same error message.

@Update：
在这种特定情况下，VC ++编译器似乎无法处理默认模板参数。为了解决这个问题，我必须将所有四个模板参数添加到typedef中，这样定义就变得像这样：

@Update: VC++ Compiler seems to be unable to process default template parameters in this particular scenario. To solve the issue I had to add all four template parameters to the typedef and so the definition became like:

template <typename K, typename V, template <typename...> class Map>
class Test {
    typedef Map<K, std::pair<V, typename std::list<K>::iterator>, std::less<K>, std::allocator<std::pair<const K, typename std::list<K>::iterator>>> MapType;
};

这就是这个问题的全部。感谢所有尝试提供帮助的人以及这位专业的绅士：我对这个问题甚至都不了解，让我们DOWNVOTE it !!!。你真的很棒！祝你最好的人。...

That would be all to this issue. Thanks for all who tried to help and for that professional gentleman with: 'I don't have even the slightest idea about this question, Let's DOWNVOTE it!!!'. You really are amazing! Wish you the best man....

推荐答案

您错过了两分。

首先， 模板模板参数 应该是这样的：

First, template template parameter should be like this:

template<参数列表>类名

所以您的

template<typename...> Map

应该是

template<typename...> class Map

第二，应该使用类型名称和依赖名称。在您的代码中， std :: list< Key> :: iterator 是一个从属名称（取决于 Key ）。因此，应改用 typename std :: list< Key> :: iterator 。

Second, you should use typename with dependent names. In your code, std::list<Key>::iterator is a dependent name (depending on Key). So, you should use typename std::list<Key>::iterator instead.

这是我正确的测试代码。

Here's my corrected test code.

#include <list>
#include <map>

template <typename Key, typename Value,
    template <typename...> class Map>
class Test
{
public:
    typedef Map<
        Key,
        std::pair<Value, typename std::list<Key>::iterator>
        > KeyToValueType;
};

int main()
{
    Test<int, char, std::map>::KeyToValueType asdf;
}

它在g ++ 4.9.1和clang ++ 3.5中均有效。

It worked both in g++ 4.9.1 and in clang++ 3.5.

这似乎是由于VC ++的愚蠢。如果将 full 模板参数提供给 std :: map （包括比较器和分配器），则可能会起作用，因为VC ++似乎无法处理在这种情况下为默认模板参数。

It seems to be due to VC++'s foolishness. It may work if you give the full template parameter to std::map, including comparer and allocator, since VC++ seems not to be able to process default template parameter in this case.

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