对函数库 std::bind 中的语法感到困惑

Question

#include <iostream>
#include <functional>
class Solver{
public:
    void print(){
        std::cout << "i solved" << std::endl;
    }
};

class ThingHandler{
public:
    void handleThing(Solver& solver){
        std::cout << "i handled something " << std::endl;
        solver.print();
    }
};

class CantHandle{
public:
    void needHelp(std::function<void(Solver&)> handleThing){
        handleThing(); // how do i here pass Solver as a parameter?
    }
};



int main() {
    ThingHandler thingHandler;
    CantHandle cantHandle;
    Solver solver;
    auto fp = std::bind(&ThingHandler::handleThing, thingHandler,solver);
    cantHandle.needHelp(fp);
    return 0;
}

在“CantHandle”类中，当我调用辅助函数时，它会要求一个参数。这是有道理的，因为被调用的函数是handleThing，它将一个对求解器的引用作为参数。但我似乎无法弄清楚 ThingHandler 类中的语法。不应该 handleThing 已经绑定了参数吗？我收到以下错误“不匹配调用‘(std::function) ()”

完整的错误信息：

错误：对‘(std::function) ()’的调用不匹配 23 |
处理事物（）； // 我如何在这里将 Solver 作为参数传递？笔记： 候选人：'_Res std::function<_res ...>::operator()(_ArgTypes ...) const [with _Res = void; _ArgTypes = {求解器&}]’ 685 |函数<_res>:: |
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ /usr/include/c++/9/bits/std_function.h:685:5：注意：候选 需要 1 个参数，提供 0 个参数 gmake[3]: * [CMakeFiles/funcObj.dir/build.make:63: CMakeFiles/funcObj.dir/main.cpp.o] 错误 1 ​​gmake[2]: 强> [CMakeFiles/Makefile2:73: CMakeFiles/funcObj.dir/all] 错误 2 gmake[1]: [CMakeFiles/Makefile2:85: CMakeFiles/funcObj.dir/rule] 错误 2 gmake: * [Makefile:118: funcObj] 错误 2

Answer 1

needHelp 方法需要一个带一个参数的函数，但您想传递一个不带参数的函数。改变这个：

void needHelp(std::function<void(Solver&)> handleThing){

到

void needHelp(std::function<void()> handleThing){

使用 lambda 而不是 bind 通常更易于编写和阅读，并且当您可以接受任何可调用时，通常不需要使用 std::function：

#include <iostream>

struct Solver{
    void print(){ 
        std::cout << "i solved" << std::endl; 
    }
};

struct ThingHandler{
    void handleThing(Solver& solver){
        std::cout << "i handled something " << std::endl;
        solver.print();
    }
};

struct CantHandle{
    template <typename F>
    void needHelp(F f){
        f();
    }
};   

int main() {
    ThingHandler thingHandler;
    CantHandle cantHandle;
    Solver solver;   
    cantHandle.needHelp(
        [&solver,&thingHandler](){
            thingHandler.handleThing(solver);
        }
    );
    return 0;
}

Answer 2

在CantHandle 类中更改您的方法并添加一个参数为Solver 类型，如下所示并传递数据（Solver 对象）：

class CantHandle{
public:
    void needHelp(std::function<void(Solver&)> handleThing, Solver data){
        handleThing(data); // how do i here pass Solver as a parameter?
    }
};

主要是在fp 之后传递solver 作为Solver 对象：

int main() {
    ThingHandler thingHandler;
    CantHandle cantHandle;
    Solver solver;
    auto fp = std::bind(&ThingHandler::handleThing, thingHandler,solver);
    cantHandle.needHelp(fp, solver);
    return 0;
}

输出将是：

i handled something
i solved

根据this link和this