函数模板、部分应用和模板参数推导

Question

我尝试让以下主要函数按预期进行编译和工作：

int main()
{
    auto square = [](int x){ return x*x; };

    typedef std::vector<int> Row;
    typedef std::vector<Row> Mat;
    Mat mat;
    auto squareElements = Curry(Map<Row>, square);
    Mat squaredMat = Map<Mat>(squareElements, mat);
}

现在我的补充代码是这样的：

#include <algorithm>
#include <functional>
#include <iterator>
#include <vector>

template <typename ContainerOut, typename ContainerIn, typename F>
ContainerOut Map( const F& f, const ContainerIn& xs )
{
    ContainerOut ys;
    // For performance reason one would use
    // ys.reserve( xs.size() )
    // and std::back_inserter instead of std::inserter
    // if ys is a std::vector.
    auto it = std::inserter( ys, end( ys ) );
    std::transform( begin( xs ), end( xs ), it, f );
    return ys;
}

template <typename Ret, typename Arg1, typename ...Args>
auto Curry( Ret f(Arg1, Args...), Arg1 arg ) -> std::function<Ret(Args...)>
{
    return [=]( Args ...args ) { return f( arg, args... ); };
}

它 does not compile.

知道如何让编译器推导出模板参数吗？

Answer 1

编译器错误说：

deduced conflicting types for parameter 'Arg1' ('const main()::<lambda(int)>&' and 'main()::<lambda(int)>')
 auto squareElements = Curry(Map<Row, decltype(square)>, square);
                                                               ^

将函数 Curry 更改为

template <typename Ret, typename Arg1, typename... Args>
auto Curry(Ret (*f)(const Arg1&, const Args&...), const Arg1& arg ) -> std::function<Ret(Args...)>
{
    return [=]( Args... args ) { return f( arg, args... ); };
}

或将函数 Map 更改为：

template <typename Container, typename F>
Container Map(F f, Container xs );

这将编译！

看看我的代码：Ideone

Answer 2

避免每次都使用 std::placeholders::_1 和 std::bind 的可能解决方案。

#include <algorithm>
#include <functional>
#include <iterator>
#include <utility>
#include <vector>

template <typename Container, typename F>
Container Map( const F& f, const Container& xs )
{
    Container ys;
    // For performance reasons one would use
    // ys.reserve( xs.size() )
    // and std::back_inserter instead of std::inserter
    // if ys is a std::vector.
    auto it = std::inserter( ys, end( ys ) );
    std::transform( begin( xs ), end( xs ), it, f );
    return ys;
}

template <typename F, typename T>
auto Curry(F&& f, T&& t)
{
    return [f = std::forward<F>(f), t = std::forward<T>(t)]
           (auto&&... args)
           { return f(t, std::forward<decltype(args)>(args)...); };
}

int main()
{
    auto square = [](int x){ return x*x; };

    typedef std::vector<int> Row;
    Row row;
    Row squaredRow = Map(square, row);

    typedef std::vector<Row> Mat;
    Mat mat;
    auto squareRow = Map<Row, decltype(square)>;
    auto squareRowElems = Curry((Map<Row, decltype(square)>), square);
    Mat squaredMat = Map(squareRowElems, mat);
}

来源：

演示：http://ideone.com/16cx0l

函数模板、部分应用和模板参数推导

function templates, partial application and template argument deduction

c++

c++11

function-templates

partial-application

type-deduction