std :: transform和std :: plus如何协同工作？ [英] How std::transform and std::plus work together?

问题描述

我正在阅读C ++参考，并通过示例遇到了std :: plus函数。这很简单，只需添加lhs和rhs。代码是：

I was reading C++ reference and came across std::plus function with an example. Which is pretty straight forward, its simply adds lhs and rhs. The code was:

#include <functional>
#include <iostream>

int main()
{
   std::string a = "Hello ";
   const char* b = "world";
   std::cout << std::plus<>{}(a, b) << '\n';
}

输出：Hello world

output: Hello world

我将其更改为

#include <functional>
#include <iostream>

int main()
{
   int a = 5;
   int b = 1;
   std::cout << std::plus<int>{}(a, b) << '\n';
}

输出：6

现在我做了

foo vector = 10 20 30 40 50
bar vector = 11 21 31 41 51

我叫：

std::transform (foo.begin(), foo.end(), bar.begin(), foo.begin(), std::plus<int>());

它给出了21 41 61 81 101，据我了解，它会将foo和bar加起来。但是如何将其传递给std :: plus函数？

and it gave 21 41 61 81 101 which I understand it is adding up both foo and bar. But how it was passed to std::plus function?

推荐答案

std :: plus<> 是functor ，对于实现 operator（）的类来说，这只是花哨的话题code>。例如：

std::plus<> is a functor, which is just fancy talk for a class that implements operator(). Here's an example:

struct plus {
    template <typename A, typename B>
    auto operator()(const A& a, const B& b) const { return a + b; }
};

std :: transform ，您大致相当于：

The std::transform you have there is roughly equivalent to:

template<typename InputIt1, typename InputIt2, 
         typename OutputIt, typename BinaryOperation>
OutputIt transform(InputIt1 first1, InputIt1 last1, InputIt2 first2, 
                   OutputIt d_first, BinaryOperation binary_op)
{
    while (first1 != last1) {
        *d_first++ = binary_op(*first1++, *first2++);
    }
    return d_first;
}

此处， binary_op 是 std :: plus<> 的名称。由于 std :: plus<> 是函子，因此C ++会将对它的调用解释为对 operator（）的调用函数，给我们我们想要的行为。

Here, binary_op is the name given to std::plus<>. Since std::plus<> is a functor, C++ will interpret the "call" to it as a call to the operator() function, giving us our desired behavior.

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