在C++ 0x中模拟finally块

Question

受到其他主题的启发,我编写了这个模拟finally块的代码:

#include <cassert>
#include <iostream>

struct base { virtual ~base(){} };

template<typename TLambda>
struct exec : base 
{
   TLambda lambda;
   exec(TLambda l) : lambda(l){}
   ~exec() { lambda(); }
};

class lambda{
    base *pbase;
public:
    template<typename TLambda>
    lambda(TLambda l): pbase(new exec<TLambda>(l)){}
    ~lambda() { delete pbase; }
};

class A{
    int a;
public:
    void start(){
        int a=1;        
        lambda finally = [&]{a=2; std::cout<<"finally executed";}; 
        try{
            assert(a==1);
            //do stuff
        }
        catch(int){
            //do stuff
        }
    }
};
int main() {
    A a;
    a.start();
}

输出(ideone):

finally executed

@Johannes似乎认为它不完全正确,并评论说:

如果编译器在复制初始化中没有删除临时文件,它可能会崩溃,因为它会使用相同的指针值删除两次

我想知道究竟是怎么回事.帮我理解问题:-)

---

编辑:

问题修复为:

class lambda{
    base *pbase;
public:
    template<typename TLambda>
    lambda(TLambda l): pbase(new exec<TLambda>(l)){}
    ~lambda() { delete pbase; }

    lambda(const lambda&)= delete;            //disable copy ctor
    lambda& operator=(const lambda&)= delete; //disable copy assignment
};

然后将其用作:

//direct initialization, no copy-initialization
lambda finally([&]{a=2;  std::cout << "finally executed" << std::endl; }); 

完整代码:http://www.ideone.com/hsX0X

Answer 1

在这个初始化中:

lambda finally = [&]{a=2; std::cout<<"finally executed";};

lambda可以使用隐式定义的复制构造函数.这将只复制原始指针pbase,然后将多次删除.

例如

$ g++ -std=c++0x -Wall -Wextra -pedantic -fno-elide-constructors lambdafun.cc 
$ ./a.out 
a.out: lambdafun.cc:29: void A::start(): Assertion `a==1' failed.
finally executedAborted (core dumped)

实际上,你的断言触发掩盖了双重删除问题,但这证明了我突出显示的崩溃.

$ g++ -std=c++0x -Wall -Wextra -pedantic -fno-elide-constructors -DNDEBUG lambdafun.cc 
$ ./a.out 
Segmentation fault (core dumped)