推断模板参数

Question

我正在尝试编写一个函数模板来推断其大部分参数以进行简单查找 table。

但是，我的努力总是出现编译器错误。这是我迄今为止最好的尝试：

// A simple templated struct that maps one key to one value
template<typename KT, typename VT>
struct LookupTable
{
    KT key;
    VT value;
};

// A lookup function that searches an array for a match.
// returns NULL if key was not found.
// returns pointer to matching value when key is found.
//
// The first parameter is a Reference to an Array of T, with a specific size N
// In this example, the size is 12.
template<typename T, size_t N, typename KT, typename VT>
VT* Find(T<KT,VT> (&A)[N], KT key)   // <== Multiple Errors on this line
{
    VT* pFoundValue = NULL;
    for (size_t i = 0; i < N; ++i)
    {
        if (key == A[i].key)
        {
            pFoundValue = &(A[i].value);
            break;
        }
    }
    return pFoundValue;
}

// Test the function with a simple example
int main(void)
{
    LookupTable<std::string, int> calendar[] = {
        { "January",   31 },
        { "February",  28 },
        { "March",     31 },
        { "April",     30 },
        { "May",       31 },
        { "June",      30 },
        { "July",      31 },
        { "August",    31 },
        { "September", 30 },
        { "October",   31 },
        { "November",  30 },
        { "December",  31 }
    };

    const int* pDays = Find(calendar, std::string("May"));

    if (pDays == NULL)
    {
        cout << "Invalid Month" << endl;
    }
    else
    {
        cout << "The month of May has " << *pDays << " Days" << endl;
    }

    _getch();
    return 0;
}

我得到的错误都在Find函数的声明上
(标有注释):

1>error C2143: syntax error : missing ')' before '<'
1>error C2143: syntax error : missing ';' before '<'
1>error C2988: unrecognizable template declaration/definition
1>error C2059: syntax error : '<'
1>error C2059: syntax error : ')'
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

看来我已经使所有参数都易于从函数调用中推断出来了。
我是否遗漏了语法错误？

Answer 1

在此：

template<typename T, size_t N, typename KT, typename VT>
VT* Find(T<KT,VT> (&A)[N], KT key)

您正在尝试使用 T 作为模板模板参数。语法是：

template<template <class, class> class T, size_t N, typename KT, typename VT>
VT* Find(T<KT,VT> (&A)[N], KT key)

Answer 2

而不是

template<typename T, size_t N, typename KT, typename VT>
VT* Find(T<KT,VT> (&A)[N], KT key)

使用

template<template <typename, typename> class T, size_t N, typename KT, typename VT>
        //^^^^^^^^^^^^^^^^^^^ Need to indicate that T is a class template
VT* Find(T<KT,VT> (&A)[N], KT key)

Answer 3

这是一个更紧凑、对 SFINAE 友好的版本，可以编译。注意我声明了数组参数const因为不需要修改

#include <type_traits>

template<typename PairT,
         typename KeyT = decltype(std::declval<PairT>().key),
         typename ValueT = decltype(std::declval<PairT>().value),
         std::size_t N>
const ValueT *
Find(const PairT (&array)[N], KeyT key)
{
  for (auto& iter : array)
    if (key == iter.key)
      return &(iter.value);
  return nullptr;
}

当然，这可以使用 STL 算法进一步简化，这也将避免为查找键创建 std::string 临时文件。

auto pred = [](const LookupTable<std::string, int>& cal){
  return cal.key == "May";
};
auto pDays = std::find_if(std::begin(calendar), std::end(calendar), pred);
if (pDays == std::end(calendar))
  std::cout << "Invalid Month" << std::endl;
else
  std::cout << "May has " << (*pDays).value << " days" << std::endl;

Answer 4

试试这个格式：

 template<size_t N, typename KT, typename VT, template<typename, typename> class T>

推断模板参数

Inferring template parameters

c++

templates

type-inference