成员属性的 std::function 模板参数推导

Question

通过这个小例子，我试图让编译器自动推导出第二个参数的模板参数。 这有效，但没有我想要的那么简洁。

struct Student {
    AgeCategory age;
    Income income;
    bool is_student;
    CreditRating credit_rating;
    bool buys_computer;
};


// This works  (A)
template<typename R>
auto calc_mean(const std::vector<Student> & in, std::function<R (Student const&)> attr)-> double
{
    const auto mean = std::accumulate(std::begin(in), std::end(in), 0.0, [&attr](auto acc, const auto& val) {
        // Call the attribute passed in
        return acc += static_cast<double>(attr(val));
    }) / static_cast<double>(in.size());
    return mean;
}

// This doesn't work (B)
template<typename T>
auto calc_mean(const std::vector<Student> & in, T attr)-> double
{
    const auto mean = std::accumulate(std::begin(in), std::end(in), 0.0, [&attr](auto acc, const auto& val) {
    // Call the attribute passed in
    return acc += static_cast<double>(attr(val));
}) / static_cast<double>(in.size());
  return mean;
}

// Caller (A) - works but I have to explicitly state the attribute type
mean_stddev<AgeCategory>(buy, &Student::age);
// Caller (B) - what I'd like to be able to do and let compiler infer types
mean_stddev(buy, &Student::age);

错误是

>..\src\Main.cpp(16): error C2672: mean_stddev': no matching overloaded function found
1>..\src\Main.cpp(16): error C2784: 'std::tuple<double,double> mean_stddev(const std::vector<Student,std::allocator<_Ty>> &,T *)': could not deduce template argument for 'T *' from AgeCategory Student::* '
1>          with
1>          [
1>              _Ty=Student
1>          ]
1>  c:\users\chowron\documents\development\projects\ml\src\Bayes.h(25): note: see declaration of mean_stddev'

我必须对 B 的函数声明做什么才能使用更简洁的语法。

Answer 1

要调用attr ，您需要使用std::invoke ：

template <class R> // <-- NB: R, not T
double calc_mean(const std::vector<Student>& in, R attr)
{
    const auto mean = std::accumulate(in.begin(), in.end(), 0.0, [&attr](auto acc, const auto& val) {
        return acc + static_cast<double>(std::invoke(attr, val));
    }) / static_cast<double>(in.size());
    return mean;
}

或者说实话：

template <class R> // <-- NB: R, not T
double calc_mean(const std::vector<Student>& in, R attr)
{
    double sum = 0.0;
    for (auto const& s : in) {
        sum += std::invoke(attr, s);
    }
    return sum / in.size();
}

invoke()是一个 C++17 函数模板，但您可以根据所附参考在 C++11 中实现它。 它将为函数、函数对象和指向成员的指针做正确的事情——这基本上是你想要的。

Answer 2

我必须对 B 的函数声明做什么才能使用更简洁的语法。

首先，您应该使用相同的模板标识符：或R或T

template<typename T>  // <--- use R here
auto calc_mean(const std::vector<Student> & in, R attr)-> double
{
    const auto mean = std::accumulate(std::begin(in), std::end(in), 0.0, [&attr](auto acc, const auto& val) {
    // Call the attribute passed in
    return acc += static_cast<double>(attr(val));
}) / static_cast<double>(in.size());
  return mean;
}