c++ - emplace_back 初始化列表错误，当初始化列表对独立变量起作用时

Question

所以我一直在制作一个对象池类，它是这样使用的:

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();
});

这很好用，但是当线程池使用的对象删除了它的复制构造函数时，例如数据成员是 std::unique_ptr 池的构造失败。在内部我使用一个 vector 来存储池:

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

然后我构建了完整的池类:

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

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

然而，这会将元素复制到 vector 中。所以我将其更改为 emplace_back 以在 vector 中就地构造对象，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});
        }
    }
} 

然而，这会出错:

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
          ]

但是我可以像这样使用初始化列表在构造函数中创建一个对象，这样编译很好。

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

我看到了这个答案，但是我无法让它工作: emplace_back not working with std::vector<std::map<int, int>>我将 std::initializer_list 的模板用作:

std::initializer_list<ObjectAndLock>

也许这是错误的？

所以我的问题是如何让 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，因为它是不可复制的:

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

Answer 1

如果您查看 emplace_back 的签名

template <class... Args>
reference emplace_back(Args&&... args);

你会发现 emplace_back 的参数类型是从你传递的参数推导出来的。 braced-init-list 可用于为特定类型的参数初始化参数。但是 {...} 本身没有类型，因此不能用于推断参数的类型。

emplace_back 所做的只是 std::forward您将传递给它的任何参数传递给元素类型的构造函数以在 vector 中就地创建元素。问题是你的

struct ObjectAndLock {
    Object object;
    bool free;
    static bool isFree(const ObjectAndLock& objectAndLock) {
        return objectAndLock.free;
    }
};

甚至没有接受参数的构造函数(隐式复制和 move 构造函数除外)。

你要做的是

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

即，为 emplace_back 初始化一个正确类型的值，以转发给隐式 move 构造函数。但这在本质上与只是做同样的事情

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

braced-init-list 与 push_back 一起工作，因为 push_back

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

确实期望元素类型的值而不是一组转发引用，因此，{...} 将最终初始化适当类型的参数...

C++20 将通过 T(…) 为聚合引入直接初始化，这将使简单的编写成为可能

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

这里。在那之前，我建议在这种情况下只使用 push_back ......

你的

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

将不可复制但也不可 move 。您声明了一个复制构造函数，这意味着不会有隐式声明的 move 构造函数 [class.copy]/8.1 .隐式声明的 move 赋值运算符的故事几乎相同 [class.copy.assign]/4 .您根本无法将不可 move 的类型作为 std::vector 的元素类型。要使 NonCopyable 可 move ，您必须定义 move 构造函数和 move 赋值运算符:

struct NonCopyable {
    std::unique_ptr<int> number = std::make_unique<int>(10);

    NonCopyable(const NonCopyable&) = delete;
    NonCopyable(NonCopyable&&) = default;
    NonCopyable& operator=(const NonCopyable&) = delete;
    NonCopyable& operator=(NonCopyable&&) = default;
};