Question

我在下面的代码中尝试打印10的乘法表。我已将函数multiply与两个参数5和2绑定在一起。我通过绑定函数来创建一个lambda。然后，我尝试将lambda传递到for_each循环以打印乘法表。我凭直觉有点知道我可能会把它推得太远了。但是我不知道确切的原因。有人可以解释。

#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;
}

我得到的错误如下所示。

/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

我正在传递一个带有两个边界和一个占位符参数的函数。

不，你不是。您的lambda看起来像这样：

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

这是您传递给std::function的可调用对象。现在，lambda只能使用一个参数而不是三个参数来调用。注意：

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

同样，f是只有一个参数的可调用对象。您不能使用三个参数来调用它，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。也许这样可以更清楚：

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；带int但不返回任何内容的可调用对象：

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

将C ++绑定函数传递给lambda

1 个答案: