在C ++中实现（类型化）K组合器 [英] Implementing the (typed) K combinator in C++

问题描述

我想从C ++中的 SK组合演算中实现K组合。 K组合器是一个高阶函数，基本上需要一些值 x ，并返回依次取值 y ，并返回 x 从它。换句话说，

  K（x）（y）== x 
  

或分步：

  intermediate = K（x）
 intermediate（y）== x 
  

c $ c> K（x）作为一个内在的东西，独立于 y 是必不可少的。此外，当简单地创建 K（x）而不调用它时，不需要指定 y 的类型 y 。一旦 K（x）（y）在代码中的某处被计算， y / p>

我试图修复我写的尝试实现K组合器的代码：

  #include< iostream> 
 
 template< class A> 
 template< class B> 
 auto K = []（A x）{
 return [=]（B y）{
 return x; 
}; 
} 
 
 int main（）
 {
 std :: cout< Hello world！\\\
; 
 auto Kx = K< int>（3）; 
 auto Kxy = Kx< float>（4.5）; 
 std :: cout<< Kxy < std :: endl; 
} 
  

输出错误：模板中的外部模板参数列表专业化或外联模板定义。我已经尝试调整模板参数，并移动它们无法使用。有人知道如何解决这个错误吗？

解决方案

Lambdas不能是模板。您可以这样做：

  #include< iostream> 
 
 auto K = []（auto x）{
 return [=]（auto y）{
 return x; 
}; 
} 
 
 int main（）
 {
 std :: cout< Hello world！\\\
; 
 auto Kx = K（3）; 
 auto Kxy = Kx（4.5）; 
 std :: cout<< Kxy < std :: endl; 
} 
  

这些被称为通用lambdas（存在自C ++ 14）基本上你想要什么。 运算符（）是每个 auto 参数的模板。

I am trying to implement the K combinator from the SK combinator calculus in C++. The K combinator is a higher-order function that basically takes some value x, and returns something which in turn takes a value y and returns x from it. In other words,

K(x)(y) == x

or step-by-step:

intermediate = K(x)
intermediate(y) == x

The ability to treat K(x) as a thing-in-itself, independent of y, is essential. Furthermore, it should not be necessary to specify the type of y when simply creating K(x) without calling it on y. The type of y can be specified once K(x)(y) is being evaluated somewhere in the code.

I am trying to fix the code I wrote which attempts to implement the K combinator:

#include <iostream>

template<class A>
template<class B>
auto K = [](A x) {
    return [=](B y) {
        return x;
    };
};

int main()
{
    std::cout << "Hello world!\n";
    auto Kx = K<int>(3);
    auto Kxy = Kx<float>(4.5);
    std::cout << Kxy << std::endl;
}

It outputs error: extraneous template parameter list in template specialization or out-of-line template definition. I have tried adjusting the template parameters and moving them around to no avail. Does anyone know how I could fix this error?

解决方案

Lambdas cannot be templates. You can do this though:

#include <iostream>

auto K = [](auto x) {
    return [=](auto y) {
        return x;
    };
};

int main()
{
    std::cout << "Hello world!\n";
    auto Kx = K(3);
    auto Kxy = Kx(4.5);
    std::cout << Kxy << std::endl;
}

These are called generic lambdas (exist since C++14), and are basically what you want. Their operator() is template for each auto parameter.

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