C++ lambda將不會調用析構函數在按值捕獲的成員上

Question

我今天在XCode下這種奇怪的lambdas行爲已經非常困難 - 在試圖追蹤iOS中圍繞代碼的幾處內存泄漏之後，我縮小了它的範圍到這一點（類似）的片段（S），我使用的是共享指針的東西的所有權分配給延期任務：

void DBStorage::dispose(std::shared_ptr<DataChunk>& dc) 
{ 
    backgroundQueue.queueTask([=]() { 
     assert(dc.use_count() == 1); 

     if (dc->isDirty()) { 
      //store to disk 
     } 
    }); 
} 

（請注意，共享指針的使用計數始終爲1，當拉姆達運行）

執行後，這個任務是空的 - 我想用pendingJob = nullptr;來調用所有按值捕獲的析構函數對象，因此是DataChunk的析構函數。 但是，它看起來像是在XCode/LLVM lc的析構函數中永遠不會被調用;使用mutable明確調用它的dtor，並用簡單的delete刪除std::function也不起作用。

這是標準行爲嗎？ 我當然可以手動調用dc.reset()，它按預期工作，但這相當重要的是使用共享指針模擬。

---

解決方案 顯然，這是一個known gcc bug。

---

的Contrib

單機與輸出採樣從Xcode中5.0.2 /鐺3.3

#include <iostream> 
#include <memory> 

void fnRef(std::shared_ptr<int>& ptr) 
{ 
    auto lambda = [=]() { std::cout << ptr.use_count() << ':' << __PRETTY_FUNCTION__ << '\n'; }; 
    lambda(); 
} 

void fnVal(std::shared_ptr<int> ptr) 
{ 
    auto lambda = [=]() { std::cout << ptr.use_count() << ':' << __PRETTY_FUNCTION__ << '\n'; }; 
    lambda(); 
} 

int main() 
{ 
    std::shared_ptr<int> ptr(new int); 
    for (int i=0; i<10; ++i) 
     fnVal(ptr); 
    std::cout << '\n'; 

    for (int i=0; i<10; ++i) 
     fnRef(ptr); 

    return 0; 
} 

LLVM/GCC輸出

3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 
3:void fnVal(std::shared_ptr<int>)::<anonymous class>::operator()() const 

2:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
3:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
4:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
5:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
6:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
7:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
8:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
9:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
10:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 
11:void fnRef(std::shared_ptr<int> &)::<anonymous class>::operator()() const 

IDEOne.com Output for same code

3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
3:fnVal(std::shared_ptr<int>)::__lambda1 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 
2:fnRef(std::shared_ptr<int>&)::__lambda0 

的Visual Studio 2013的輸出

3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 
3:fnVal::<lambda_67137a3f93ee478c018cc7068004c9fd>::operator() 

2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 
2:fnRef::<lambda_70f241d4201227663d23c74be170d302>::operator() 

Answer 1

正如注意到@DaveS這可能是一個known gcc bug - 捕獲引用存儲作爲參考。

使用存儲的lambda時，一個好的經驗法則是避免=，因爲應該小心處理存儲狀態。

void DBStorage::dispose(std::shared_ptr<DataChunk>& dc) 
{ 
    std::shared_ptr<DataChunk> data_to_store = dc; 
    backgroundQueue.queueTask([data_to_store]() { // maybe add `,this` to the capture list 
    assert(data_to_store.use_count() == 1); 
    if (data_to_store->isDirty()) { 
     //store to disk 
    } 
    }); 
} 

或：

void DBStorage::dispose(std::shared_ptr<DataChunk> data_to_store) 
{ 
    backgroundQueue.queueTask([data_to_store]() { // maybe add `,this` to the capture list 
    assert(data_to_store.use_count() == 1); 
    if (data_to_store->isDirty()) { 
     //store to disk 
    } 
    }); 
} 

以主動建議的第二位，std::function s爲不lambda表達式，並調用一個theLambda是一種誤導。