如何在 C++ 中进行柯里化?

什么是咖喱?

如何在 C++ 中实现柯里化?

请解释 STL 容器中的绑定器?

解决方案

简而言之，柯里化需要一个函数 f(x, y) 并给定一个固定的 Y，给出一个新函数 g(x) where

g(x) == f(x, Y)

这个新函数可以在只提供一个参数的情况下被调用，并将调用传递给带有固定 Y 参数的原始 f 函数.>

STL 中的绑定器允许您为 C++ 函数执行此操作.例如:

#include #include #include <向量>使用命名空间标准；//声明一个二元函数对象类加法器:public binary_function{上市:int operator()(int x, int y) const{返回 x + y;}};int main(){//初始化一些样本数据向量一、乙；a.push_back(1);a.push_back(2);a.push_back(3);//这里我们声明了一个函数对象 f 并尝试一下加法器 f;cout<<"f(2, 3) = " <<f(2, 3)＜＜结束;//transform() 需要一个带一个参数的函数，所以我们使用//bind2nd 创建一个基于 f 的新函数，它需要一个//参数并将其加 5变换(a.begin(), a.end(), back_inserter(b), bind2nd(f, 5));//输出 b 看看我们得到了什么cout<<"b = [" <<结束;for (vector::iterator i = b.begin(); i != b.end(); ++i) {cout<<" " <<*我＜＜结束;}cout<<"]" <<结束;返回0；}

What is currying?

How can currying be done in C++?

Please Explain binders in STL container?

解决方案

In short, currying takes a function f(x, y) and given a fixed Y, gives a new function g(x) where

g(x) == f(x, Y)

This new function may be called in situations where only one argument is supplied, and passes the call on to the original f function with the fixed Y argument.

The binders in the STL allow you to do this for C++ functions. For example:

#include <functional>
#include <iostream>
#include <vector>

using namespace std;

// declare a binary function object
class adder: public binary_function<int, int, int> {
public:
    int operator()(int x, int y) const
    {
        return x + y;
    }
};

int main()
{
    // initialise some sample data
    vector<int> a, b;
    a.push_back(1);
    a.push_back(2);
    a.push_back(3);

    // here we declare a function object f and try it out
    adder f;
    cout << "f(2, 3) = " << f(2, 3) << endl;

    // transform() expects a function with one argument, so we use
    // bind2nd to make a new function based on f, that takes one
    // argument and adds 5 to it
    transform(a.begin(), a.end(), back_inserter(b), bind2nd(f, 5));

    // output b to see what we got
    cout << "b = [" << endl;
    for (vector<int>::iterator i = b.begin(); i != b.end(); ++i) {
        cout << "  " << *i << endl;
    }
    cout << "]" << endl;

    return 0;
}