如何使用 lambdas 和函数作为 unique_ptr 的自定义删除器

Question

我正在测试 unique_ptr 的自定义删除器。 奇怪的是，只有作为函数对象的删除器才能正常工作。 如果我用函数或 lambda 替换它们，程序将无法编译。 我做错了什么？

这是我完整的测试程序

#include <iostream>
#include <memory>

using namespace std;


class Vehicle {
public:
    Vehicle(){ cout<<"Vehicle constructor..."<<endl;}
    virtual ~Vehicle(){cout<<"~Vehicle destructor..."<<endl;}
    virtual void go()=0;
};

class Car:public Vehicle {
    public:
    void go() override {
        cout<<"Going by car..."<<endl;
    }
};

class Bicycle:public Vehicle {
    public:
    void go() override {
        cout<<"Going by bicycle..."<<endl;
    }
};

// Custom deleters
auto CustomLambdaDeleter = [](Vehicle* v){
    cout<<"Custom lambda deleter called..."<<endl;
    delete v;
};

void CustomFunctionDeleter(Vehicle* v){
    cout<<"Custom function deleter called..."<<endl;
    delete v;
}

struct CustomFunctorDeleter
{
    void operator()(Vehicle* v ) const {
        cout<<"Custom functor deleter called..."<<endl;
        delete v;
    }
};


// Doesn't compile
//using VehiclePtr = unique_ptr<Vehicle, decltype(CustomLambdaDeleter)>;
// Doesn't compile
//using VehiclePtr = unique_ptr<Vehicle, decltype(&CustomFunctionDeleter)>;
// Works ok
using VehiclePtr = unique_ptr<Vehicle, CustomFunctorDeleter>;


class VehicleFactory {
public:
    static VehiclePtr createVehicle(string type){
        VehiclePtr vptr;
        if("bicycle"==type) {
            vptr.reset(new Bicycle());
            // This also works
            // vptr= (VehiclePtr) new Bicycle();
            return vptr;
        }
        else if("car"==type) {

           vptr.reset( new Car());
           return vptr;
        }

        return nullptr;
    }
};


void vehicleFactoryTest(){
    cout<<"* Starting vehicleFactoryTest()..."<<endl;
    auto firstVehicle = VehicleFactory::createVehicle("bicycle");
    firstVehicle->go();
    auto newCar = VehicleFactory::createVehicle("car");
    newCar->go();
}

int main(int, char **)
{

    vehicleFactoryTest();
    return 0;
}

Answer 1

问题不在于任何一个

using VehiclePtr = unique_ptr<Vehicle, decltype(CustomLambdaDeleter)>;

或

using VehiclePtr = unique_ptr<Vehicle, CustomFunctorDeleter>;

这两个是自己编译的。问题在于createVehicle中的以下行

VehiclePtr vptr;

这里，您默认构造一个unique_ptr，它不会在使用 lambda 删除器的情况下编译，因为 lambda 不是默认可构造的。所以你需要

VehiclePtr vptr{nullptr, CustomLambdaDeleter};

在您使用函数指针的情况下，您尝试使用nullptr 删除器默认构造unique_ptr，这是不允许的。修复方法类似，在这种情况下您需要将指针传递给函数。

VehiclePtr vptr{nullptr, CustomFunctionDeleter};

您在createVehicle 的最终返回语句中也犯了类似的错误。将该行更改为

return vptr;