传递给 lambda 的 C++ 绑定函数

Question

我在下面的代码中尝试打印 10 的乘法 table。我用两个参数 5 和 2 绑定了函数 multiply。我正在传递绑定函数来创建一个 lambda。然后我试图将 lambda 传递给 for_each 循环以打印乘法 Table。我凭直觉知道我可能有点过头了。但我不知道确切的原因。谁能解释一下。

#include <iostream>
#include <vector>
#include <algorithm>
#include <thread>
#include <functional>
#include <future>
#include <array>
#include <unistd.h>

using namespace std;
using namespace std::placeholders;


int multiply(int a, int b, int c)
{
    return a*b*c;
}

int main()
{
    auto f = std::bind(multiply, 5, 2, _1);

    std::function<int(int,int,int)> f1 = [f](int a){cout << "Multiplication Table (10) :" << f(a) << endl; };

    vector<int> vec = {1,2,3,4,5,6,7,8,9,10};

    for_each(vec.begin(), vec.end(), f1);

    return 0;
}

我遇到的错误如下所示。

/home/karthik/Workspace/cpppen/learning/main.cpp: In function ‘int main()’:
/home/karthik/Workspace/cpppen/learning/main.cpp:26:107: error: conversion from ‘main()::<lambda(int)>’ to non-scalar type ‘std::function<int(int, int, int)>’ requested
     std::function<int(int,int,int)> f1 = [f](int a){cout << "Multiplication Table (10) :" << f(a) << endl;};
                                                                                                           ^
In file included from /usr/include/c++/7/algorithm:62:0,
                 from /home/karthik/Workspace/cpppen/learning/main.cpp:6:
/usr/include/c++/7/bits/stl_algo.h: In instantiation of ‘_Funct std::for_each(_IIter, _IIter, _Funct) [with _IIter = __gnu_cxx::__normal_iterator<int*, std::vector<int> >; _Funct = std::function<int(int, int, int)>]’:
/home/karthik/Workspace/cpppen/learning/main.cpp:30:40:   required from here
/usr/include/c++/7/bits/stl_algo.h:3884:5: error: no match for call to ‘(std::function<int(int, int, int)>) (int&)’
  __f(*__first);
  ~~~^~~~~~~~~~
In file included from /usr/include/c++/7/functional:58:0,
                 from /home/karthik/Workspace/cpppen/learning/main.cpp:8:
/usr/include/c++/7/bits/std_function.h:701:5: note: candidate: _Res std::function<_Res(_ArgTypes ...)>::operator()(_ArgTypes ...) const [with _Res = int; _ArgTypes = {int, int, int}]
     function<_Res(_ArgTypes...)>::
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 1

I am passing a function with two bound and one placeholder parameter.

不，你不是。您的 lambda 看起来像这样：

[f](int a) {
    cout << "Multiplication Table (10) :" << f(a) << endl;
}

这是您要传递给 std::function 的可调用对象。现在，lambda 只能用一个参数调用，而不是三个。注意：

[/*...*/](int a){ /*...*/ }
//        ^^^^^
//      one parameter

同样，f是一个只有一个参数的可调用对象。您不能使用三个参数调用它，因为您已将两个参数绑定到特定值，因此，就所有意图和目的而言，没有三个参数。也许这更清楚：

auto add = [](int a, int b) { return a + b; };
auto addTo5 = [&add](int a) { return add(a, 5); };

add(1, 2); // ok, lambda takes two parameters
addTo5(3); // ok, lambda takes one parameter
addTo5(1, 2); // not ok, lambda doesn't take two parameters

std::function<int(int, int)> fadd = add; // ok
std::function<int(int)> faddTo5 = addTo5; // ok
std::function<int(int, int)> faddTo5fail = addTo5; // not ok, same reason

// addTo5 is approximately what std::bind does: It generates an object which has
// several variables "fixed", and so only takes the placeholder arguments that aren't
// specified.

因此，解决方法是更改 f1 的类型以反映您实际存储的内容；一个不需要 int 和 returns 的可调用对象：

std::function<void(int)> f1 = /*...*/;
//            ^^^^
//          lambda returns nothing

传递给 lambda 的 C++ 绑定函数

C++ bound function passed to a lambda

c++

lambda

std-function