emplace_back初始化列表错误,当初始化列表在独立变量上工作时

emplace_back initialisation list error, when initialisation list works on standalone variable

本文关键字:列表 初始化 变量 独立 工作 back 错误 emplace 当初      更新时间:2023-10-16

所以我一直在做一个对象池类,它是这样用的:

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();
});

这工作正常,但是当线程池使用的对象删除其复制构造函数时,例如数据成员是池构造失败std::unique_ptr。 在内部,我使用一个向量来存储池:

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

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}) 
{}

这将使用此处列出的第三个重载构造向量 https://en.cppreference.com/w/cpp/container/vector/vector

但是,这会将元素复制到向量中。 所以我把它改成emplace_back,在向量中构造对象,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,这会导致以下错误:

ProjectsObjectPoolpublic_includeObjectPoolObjectPool.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 Studio2019CommunityVCToolsMSVC14.23.28105includevector(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,我可以使用初始值设定项列表在构造函数中创建一个对象,如下所示,它可以很好地编译。

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

我已经看到了这个答案,但是我无法让它工作: emplace_back不适用于 std::vector<std::map><int,>> 我使用std::initializer_list的诱惑作为:

std::initializer_list<ObjectAndLock>

也许这是错误的?

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

下面是一个失败的示例类,因为它不可复制:

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

如果你看一下emplace_back的签名

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

您会发现emplace_back参数的类型是从您传递的参数中推导出来的。大括号初始化列表可用于初始化特定类型参数的参数。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传递给它的任何参数传递给元素类型的构造函数,以在向量中就地创建元素。问题在于您的

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(隐式复制和移动构造函数除外(。

您所要做的就是

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

即,初始化正确类型的值,以便emplace_back转发到隐式移动构造函数。但这本质上与仅仅做一回事

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

大括号初始化列表适用于push_back,因为push_back

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

确实需要元素类型的值而不是一包转发引用,因此,{…}最终将初始化适当类型的参数......

C++20 将通过聚合的T(…)引入直接初始化

,这使得简单地写入成为可能
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......

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,但也不能移动。您声明了一个复制构造函数,这意味着不会有隐式声明的移动构造函数 [class.copy]/8.1。对于隐式声明的移动赋值运算符 [class.copy.assign]/4,情况几乎相同。您根本不能将不移动类型作为std::vector的元素类型。要使NonCopyable可移动,您必须定义一个移动构造函数和移动赋值运算符:

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;
};