使用依赖于非类型模板参数的泛型函数的签名作为模板模板参数 [英] use the signature of a generic function dependent on a non-type template argument as template template argument

问题描述

下面的小程序可以编译并运行，它允许将类型为unsigned的运行时变量index与一组具有一个类型为unsigned的模板参数J的模板函数进行桥接.

The small program below, which compiles and runs, allows to bridge a runtime variable index of type unsigned with a set of template functions having one template argument J of type unsigned.

如果需要进一步说明，请在此更好地解释问题.

In case further clarifications are needed, this is better explained in this question.

我编写的辅助函数使用模板模板参数，以从原始函数中推断出尽可能多的信息.问题是，除了创建两个包装器wrap_foo和wrap_zoo之外，我找不到定义模板模板参数FunWrap的更好方法，而我想直接将foo和zoo.有办法吗?

The auxiliary functions I wrote use a template template argument, to infer as much info as possible from the original function. The problem is that I could not find a better way to define the template template argument FunWrap other than creating the 2 wrappers wrap_foo and wrap_zoo, whereas I would have liked to use as template template argument directly foo and zoo. Is there a way to do that?

#include <iostream>
#include <utility>
#include <array>
using namespace std;

// boiler plate stuff

template <template <unsigned J> typename FunWrap, unsigned... Is>
decltype(auto) get_fun_ptr_aux(std::integer_sequence<unsigned, Is...>, unsigned i)
{
    typedef decltype(&FunWrap<1>::run) FunPtr;
    constexpr static std::array<FunPtr, sizeof...(Is)> fun_ptrs = { &FunWrap<Is>::run... };
    return fun_ptrs[i];
}

template <template <unsigned J> typename FunWrap, unsigned N>
decltype(auto) get_fun_ptr(unsigned i)
{
    return get_fun_ptr_aux<FunWrap>(std::make_integer_sequence<unsigned, N>{}, i);
}

// template functions to be bridged with runtime arguments
// two functions with same template arguments but different signature

template <unsigned J>
void foo() {  cout << J << "\n"; }

template <unsigned J>
double zoo(double x) { return x + J; }

// 1 wrapper per each function

template <unsigned J>
struct wrap_foo {
    static void *run() { foo<J>(); }  // same signature as foo
};

template <unsigned J>
struct wrap_zoo {
    static double run(double x) { return zoo<J>(x); } // same signature as zoo
};

int main()
{
    unsigned index = 5;
    (*get_fun_ptr<wrap_foo,10>(index))();
    cout << (*get_fun_ptr<wrap_zoo,10>(index))(3.5) << "\n";
    return 0;
}

推荐答案

我想直接将foo和zoo用作模板模板参数.有办法吗?

I would have liked to use as template template argument directly foo and zoo. Is there a way to do that?

由于foo和zoo是模板函数，因此恐怕答案是否定的.众所周知，C ++在处理重载集/模板方面很糟糕.示例:将歧义重载的成员函数指针作为模板参数传递

As foo and zoo are template functions, I'm afraid the answer is no. C++ is notoriously bad at dealing with overload sets/templates. Example: Disambiguate overloaded member function pointer being passed as template parameter

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