thread_local std :: unique_ptr释放不调用析构函数 [英] thread_local std::unique_ptr release not calling destructor

问题描述

为什么在此代码中调用析构函数：

  #include< iostream> 
 #include< thread> 
 #include< memory> 
 
 class base {
 public：
 base（）{
 std :: cout< base（）< std :: endl; 
} 
 virtual〜base（）{
 std :: cout< 〜base（）<< std :: endl; 
} 
 base（const base&）= delete; 
 base（base&&）= delete; 
 base& operator =（const base&）= delete; 
 base& operator =（base&&）= delete; 
}; 
 
类派生final：public base {
 public：
 derived（）{
 std :: cout< derived（）<< std :: endl; 
} 
 virtual〜derived（）{
 std :: cout< 〜derived（）<< std :: endl; 
} 
}; 
 
 
 void foo（）{
 static thread_local std :: unique_ptr< base> ptr; 
 if（！ptr）{
 std :: cout<< new ptr：< std :: this_thread :: get_id（）<< std :: endl; 
 ptr.reset（new derived（））; 
} else {
 std :: cout<< release ptr：< std :: this_thread :: get_id（）<< std :: endl; 
 ptr.release（）; //我期望析构函数在这里被调用？ 
} 
} 
 
 void thread_main（）{
 foo（）; 
 foo（）; 
} 
 
 int main（）
 {
 std :: thread thread1（thread_main）; 
 thread1.join（）; 
 return 0; 
} 
  

输出：

  new ptr：140671459997440 
 base（）
 derived（）
 release ptr：140671459997440 
  

我会期望：

  new ptr：140671459997440 
 base（）
 derived（）
 release ptr：140671459997440 
〜derived（）
〜base（）
  

使用 gcc 4.9.1

解决方案

使用 ptr.reset（）; 替换 ptr.release（）; p>

Why isn't the destructor called in this code:

#include <iostream>
#include <thread>
#include <memory>

class base {
    public:
        base() {
            std::cout << "base()" << std::endl;
        }
        virtual ~base() {
            std::cout << "~base()" << std::endl;
        }
        base(const base&) = delete;
        base(base&&) = delete;
        base& operator=(const base&) = delete;
        base& operator=(base&&) = delete;
};

class derived final : public base {
    public:
        derived() {
            std::cout << "derived()" << std::endl;
        }
        virtual ~derived() {
            std::cout << "~derived()" << std::endl;
        }
};


void foo() {
    static thread_local std::unique_ptr<base> ptr;
    if (!ptr) {
        std::cout << "new ptr:" << std::this_thread::get_id() << std::endl;
        ptr.reset(new derived());
    } else {
        std::cout << "release ptr:" << std::this_thread::get_id() << std::endl;
        ptr.release();  // I would expect the destructor to be called here?!
    }
}

void thread_main() {
    foo();
    foo();
}

int main()
{
    std::thread thread1(thread_main);
    thread1.join();
    return 0;
}

Output:

new ptr:140671459997440
base()
derived()
release ptr:140671459997440

I would expect:

new ptr:140671459997440
base()
derived()
release ptr:140671459997440
~derived()
~base()

Using gcc 4.9.1

解决方案

Replace ptr.release(); with ptr.reset();.

这篇关于thread_local std :: unique_ptr释放不调用析构函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！