如果您避免“裸”资源(如运算符新,裸指针,裸互斥体等),并将所有内容都包含在具有适当RAII行为的容器或类中,那么即使存在异常,也不会有您所描述的问题.

也就是说,不要在构造函数中获取裸体资源.而是创建一个本身跟随RAII的对象的实例.这样,即使您的构造函数失败(即创建实例的构造函数),将调用初始化对象的析构函数.

所以,这是不好的做法：

#include<iostream>
#include<stdexcept>

struct Bad {
  Bad() {
    double *x = new double;
    throw(std::runtime_error("the exception was thrown"));
  }

  ~Bad() {
    delete x;
    std::cout<<"My destructor was called"<<std::endl;
  }

  double *x;  
};

int main() {
  try {
    Bad bad;
  } catch (const std::exception &e) {
    std::cout<<"We have a leak! Let's keep going!"<<std::endl;
  }
  std::cout<<"Here I am... with a leak..."<<std::endl;
  return 0;
}@H_301_13@ 
 
输出：
 
  
 
We have a leak! Let's keep going!
Here I am... with a leak...@H_301_13@ 
 
与这个愚蠢和愚蠢的良好实施相比：
 
  
 
#include<iostream>
#include<stdexcept>

struct Resource {

  Resource() {
    std::cout<<"Resource acquired"<<std::endl;    
  }

  ~Resource() {
    std::cout<<"Resource cleaned up"<<std::endl;        
  }

};

struct Good {
  Good() {
    std::cout<<"Acquiring resource"<<std::endl;
    Resource r;
    throw(std::runtime_error("the exception was thrown"));
  }

  ~Good() {
    std::cout<<"My destructor was called"<<std::endl;
  }  
};


int main() {
  try {
    Good good;
  } catch (const std::exception &e) {
    std::cout<<"We DO NOT have a leak! Let's keep going!"<<std::endl;
  }
  std::cout<<"Here I am... without a leak..."<<std::endl;
  return 0;
}@H_301_13@ 
 
输出：
 
  
 
Acquiring resource
Resource acquired
Resource cleaned up
We DO NOT have a leak! Let's keep going!
Here I am... without a leak...@H_301_13@ 
 
我的观点如下：尝试将需要解放的所有资源封装到构造函数不抛出的类中,析构函数正确地释放资源.然后,在析构函数可能抛出的其他类中,只需创建包装资源的实例,并将保证获取的资源包装器的析构函数将被清理.
 
 
以下可能是一个更好的例子：
 
  
 
#include<mutex>
#include<iostream>
#include<stdexcept>

// a program-wide mutex
std::mutex TheMutex;

struct Bad {
  Bad() {
    std::cout<<"Attempting to get the mutex"<<std::endl;
    TheMutex.lock();
    std::cout<<"Got it! I'll give it to you in a second..."<<std::endl;
    throw(std::runtime_error("Ooops,I threw!"));
    // will never get here...
    TheMutex.unlock();
    std::cout<<"There you go! I released the mutex!"<<std::endl;    
  }  
};

struct ScopedLock {
  ScopedLock(std::mutex& mutex)
      :m_mutex(&mutex) {
    std::cout<<"Attempting to get the mutex"<<std::endl;
    m_mutex->lock();
    std::cout<<"Got it! I'll give it to you in a second..."<<std::endl;    
  }

  ~ScopedLock() {
    m_mutex->unlock();
    std::cout<<"There you go! I released the mutex!"<<std::endl;        
  }
  std::mutex* m_mutex;      
};

struct Good {
  Good() {
    ScopedLock autorelease(TheMutex);
    throw(std::runtime_error("Ooops,I threw!"));
    // will never get here
  }  
};


int main() {
  std::cout<<"Create a Good instance"<<std::endl;
  try {
    Good g;
  } catch (const std::exception& e) {
    std::cout<<e.what()<<std::endl;
  }

  std::cout<<"Now,let's create a Bad instance"<<std::endl;
  try {
    Bad b;
  } catch (const std::exception& e) {
    std::cout<<e.what()<<std::endl;
  }

  std::cout<<"Now,let's create a whatever instance"<<std::endl;
  try {
    Good g;
  } catch (const std::exception& e) {
    std::cout<<e.what()<<std::endl;
  }

  std::cout<<"I am here despite the deadlock..."<<std::endl;  
  return 0;
}@H_301_13@ 
 
输出(用gcc 4.8.1编译,使用-std = c 11)：
 
  
 
Create a Good instance
Attempting to get the mutex
Got it! I'll give it to you in a second...
There you go! I released the mutex!
Ooops,I threw!
Now,let's create a Bad instance
Attempting to get the mutex
Got it! I'll give it to you in a second...
Ooops,let's create a whatever instance
Attempting to get the mutex@H_301_13@ 
 
现在,请不要按照我的例子,并创建自己的范围守卫. C(特别是C 11)设计了RAII,并提供了丰富的终身经理.例如,std :: fstream将自动关闭,[std :: lock_guard] [2]将在我的示例中尝试做,std :: unique_ptr或std :: shared_ptr将会处理破坏.
 
 
最好的建议？阅读RAII(并根据它设计),使用标准库,不要创建裸露的资源,并熟悉Herb Sutter关于“异常安全”所说的话(请继续阅读他的website或谷歌“Herb Sutter Exception Safety”)