emplace_back初始化列表错误，当初始化列表适用于独立变量时

问题描述 投票：0回答：1

所以我一直在制作对象池类，它的用法如下：

class MagicTrick {
public:
    MagicTrick(int magic) : _magic(magic)
    {}
    int predict() {
        return _number * _magic;
    }
private:
    int _magic;
    int _number = 2;
};
const std::size_t poolSize = 1;
ObjectPool<MagicTrick> magicTrickPool(poolSize, 5);

const int number = magicTrickPool.schedule([](MagicTrick& magicTrick){
    return magicTrick.predict();
});

这可以很好地工作，但是当线程池使用的对象的副本构造函数被删除时，例如数据成员为std::unique_ptr，则池的构造失败。在内部，我使用向量存储池：

struct ObjectAndLock {
    Object object;
    bool free;
    static bool isFree(const ObjectAndLock& objectAndLock) {
        return objectAndLock.free;
    }
};
std::vector<ObjectAndLock> objectAndLocks;

并且我构造了完整的池类：

template<typename Object> 
class ObjectPool {
    template<typename ...Args>
    ObjectPool(std::size_t poolSize, Args&&... objectArgs) 
        : objectAndLocks(poolSize, { {std::forward<Args>(objectArgs)...}, true}) 
{}

这将构造具有此处列出的第三个重载的向量https://en.cppreference.com/w/cpp/container/vector/vector

但是这会将元素复制到向量中。因此，我将其更改为emplace_back以在矢量中适当地构造对象，如下所示：

template<typename Object> 
class ObjectPool {
    template<typename ...Args>
    ObjectPool(std::size_t poolSize, Args&&... objectArgs)
    {
        if(poolSize == 0){
            throw std::runtime_error("poolSize must be greater than 0");
        }
        objectAndLocks.reserve(poolSize);
        for (std::size_t i = 0; i < poolSize; i++)
        {
            objectAndLocks.emplace_back({Object{std::forward<Args>(objectArgs)...}, true});
        }
    }
}

但是这有错误：

Projects\ObjectPool\public_include\ObjectPool\ObjectPool.hpp(87): error C2660: 'std::vector<object_pool::ObjectPool<MagicTrick>::ObjectAndLock,std::allocator<_Ty>>::emplace_back': function does not take 1 arguments
          with
          [
              _Ty=object_pool::ObjectPool<MagicTrick>::ObjectAndLock
          ]
  C:\Program Files (x86)\Microsoft Visual Studio\2019\Community\VC\Tools\MSVC\14.23.28105\include\vector(651): note: see declaration of 'std::vector<object_pool::ObjectPool<MagicTrick>::ObjectAndLock,std::allocator<_Ty>>::emplace_back'
          with
          [
              _Ty=object_pool::ObjectPool<MagicTrick>::ObjectAndLock
          ]

但是我可以使用初始化列表在构造函数中创建一个对象，这样可以很好地进行编译。

ObjectAndLock hello = { Object{std::forward<Args>(objectArgs)...}, true };

我已经看到了这个答案，但是我无法使它起作用：emplace_back not working with std::vector<std::map<int, int>>我将std::initializer_list的模板设为：

std::initializer_list<ObjectAndLock>

也许这是错误的？

所以我的问题是如何使emplace_back正常工作？我最多可以使用c ++ 17

这是一个失败的示例类：

struct NonCopyable {
    std::unique_ptr<int> number = std::make_unique<int>(10);
    NonCopyable(const NonCopyable& other) = delete;
    NonCopyable& operator=(const NonCopyable& other) = delete;
};

为了完整起见，这里是完整的类：

#ifndef OBJECTPOOL_H
#define OBJECTPOOL_H
#include <vector>
#include <functional>
#include <map>
#include <mutex>
#include <condition_variable>
#include <type_traits>
#include <algorithm>
#include <stdexcept>
#include <exception>
namespace object_pool {
    namespace internal {
        template <typename Function>
        class DeferToDestruction {
            Function _function;
        public:
            DeferToDestruction(Function function) : _function(function) {}
            ~DeferToDestruction() { _function(); }
        };
    }
    template<typename Object> 
    class ObjectPool {
    public:
        /*!
        @brief Create an object pool for 

        @param poolSize - Size of object pool, this must be atleast 1
        @param objectArgs... - Arguments to construct the objects in the pool

        Complete Example:
        @code
            class MagicTrick {
            public:
                MagicTrick(int magic) : _magic(magic)
                {}
                int predict() {
                    return _number * _magic;
                }
            private:
                int _magic;
                int _number = 2;
            };

            std::size_t poolSize = 5;
            object_pool::ObjectPool<MagicTrick> magicTrickPool(poolSize, 5);

            const int number = magicTrickPool.schedule([](MagicTrick& magicTrick){
                return magicTrick.predict();
            });
        @endcode

        Zero Argument Constructor Example:
        @code            
            struct ZeroArgs {
                int number = 2;
            };

            object_pool::ObjectPool<ZeroArgs> zeroArgsPool(1);
        @endcode

        Multiple Argument Constructor Example:
        @code
            class MultiArgs {
            public:
                MultiArgs(std::string name, int age, bool alive) {
                   _number = name.size() + age + (alive ? 5 : -5);
                }
                int predict() {
                    return _number * 2;
                }
            private:
                int _number = 2;
            };

            object_pool::ObjectPool<MultiArgs> multiArgsPool(1, "bob", 99, true);
        @endcode
        */
        template<typename ...Args>
        ObjectPool(std::size_t poolSize, Args&&... objectArgs)
        {
            if(poolSize == 0){
                throw std::runtime_error("poolSize must be greater than 0");
            }
            objectAndLocks.reserve(poolSize);
            for (std::size_t i = 0; i < poolSize; i++)
            {
                objectAndLocks.emplace_back({Object{std::forward<Args>(objectArgs)...}, true});
            }
        }

        ~ObjectPool(){ 

            std::unique_lock<std::mutex> lock(objectAndLocksMutex);            
            const auto allobjectAndLocksFree = [this]() {
                return std::all_of(std::begin(objectAndLocks), std::end(objectAndLocks), ObjectAndLock::isFree);
            };
            if(allobjectAndLocksFree()) {
                return;
            }
            conditionVariable.wait(lock, allobjectAndLocksFree);
        }

        /*!
        @brief Schedule access to the pool

        @param callback - An callable with the the argument being a reference to the class stored in the object pool.

        @return Returns return from the callback function, including void

        Simple Example:
        @code
            const int number = magicTrickPool.schedule([](MagicTrick& magicTrick){
                return magicTrick.predict();
            });
        @endcode
        */
        template<typename FunctionWithObjectAsParameter>
        auto schedule(FunctionWithObjectAsParameter&& callback)
        {
            const auto findFreeObject = [this]() {
                return std::find_if(std::begin(objectAndLocks), std::end(objectAndLocks), ObjectAndLock::isFree);
            };

            std::unique_lock<std::mutex> lock(objectAndLocksMutex);
            auto freeObject = findFreeObject();
            if(freeObject == std::end(objectAndLocks)) {
                conditionVariable.wait(lock, [this, &freeObject, &findFreeObject]{
                    freeObject = findFreeObject();
                    return freeObject != std::end(objectAndLocks);
                });
            }
            freeObject->free = false;
            lock.unlock();
            internal::DeferToDestruction freeAndUnlockAndNotify([this, &freeObject] () {
                {
                    std::scoped_lock<std::mutex> lock(objectAndLocksMutex);
                    freeObject->free = true;
                }
                conditionVariable.notify_one();
            });
            return callback(freeObject->object);
        }
    private:    
        struct ObjectAndLock {
            Object object;
            bool free;
            static bool isFree(const ObjectAndLock& objectAndLock) {
                return objectAndLock.free;
            }        
        };
        std::vector<ObjectAndLock> objectAndLocks;
        std::mutex objectAndLocksMutex;
        std::condition_variable conditionVariable;
    };
}
#endif
c++ templates c++17 move-semantics emplace
1个回答
0
投票

如果您查看emplace_back的签名

emplace_back

您会发现template <class... Args> reference emplace_back(Args&&... args); 的参数类型是根据您传递的参数推导出来的。 braced-init-list可以用于为特定类型的参数初始化参数。但是emplace_back本身没有类型，因此不能用于从中推导参数的类型。

{…}所做的就是简单地将emplace_back传递给元素类型的构造函数以在向量中就地创建元素的任何参数。问题在于您的

std::forward

甚至没有带参数的构造器（隐式复制和移动构造器除外）。

您需要做的是

std::forward

即为struct ObjectAndLock { Object object; bool free; static bool isFree(const ObjectAndLock& objectAndLock) { return objectAndLock.free; } }; 初始化一个正确类型的值，以转发到隐式move构造函数。但这基本上和只做同一件事

objectAndLocks.emplace_back(ObjectAndLock{Object{std::forward<Args>(objectArgs)...}, true});

braced-init-list与emplace_back一起使用是因为objectAndLocks.push_back({Object{std::forward<Args>(objectArgs)...}, true}); 

push_back

期望元素类型的值而不是一堆转发引用，因此，push_back将最终初始化适当类型的参数...

C ++ 20将为聚合引入push_back的初始化语法，这使得可以简单地编写]]

void push_back(const T& value);
void push_back(T&& value);

这里。在此之前，我建议在这种情况下只使用{…} ...
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