c ++:std :: tr1 :: shared_ptr来自此

Question

我有以下代码:

#include <memory>

class Foo;
typedef std::tr1::shared_ptr<Foo> pFoo_t;

class DoSomething
{
public:
    static void doSomething( pFoo_t p) { printf( "doing something...\n"); }
    static void doSomethingElse( pFoo_t p) { printf( "doing something else...\n"); }
};

class Foo
{
public:
    Foo() { printf( "foo()\n"); }
    ~Foo() { printf( "~foo()\n"); }
public:
    void doSomething() { DoSomething::doSomething(pFoo_t(this)); }
    void doSomethingElse() { DoSomething::doSomethingElse(pFoo_t(this)); }
};

int _tmain(int argc, _TCHAR* argv[])
{
    Foo foo;
    foo.doSomething();
    foo.doSomethingElse();

    return 0;
}

我开始这个样本,然后我得到下一个断言:_BLOCK_TYPE_IS_VALID(pHead-> nBloakUse).

我怎么能避免这个？

我使用以下代码来解决此问题:

class Foo;
typedef std::tr1::shared_ptr<Foo> pFoo_t;

class DoSomething
{
public:
    static void doSomething( pFoo_t p) { printf( "doing something...\n"); }
    static void doSomethingElse( pFoo_t p) { printf( "doing something else...\n"); }
};

class Foo
{
public:
    void Init(pFoo_t _pFoo) { m_pFoo = _pFoo; }
    Foo() { printf( "foo()\n"); }
    ~Foo() { printf( "~foo()\n"); }
public:
    void doSomething() { DoSomething::doSomething(m_pFoo.lock()); }
    void doSomethingElse() { DoSomething::doSomethingElse(m_pFoo.lock()); }
private:
    std::tr1::weak_ptr<Foo> m_pFoo;
};

int _tmain(int argc, _TCHAR* argv[])
{
    {
        Foo * foo = new Foo();
        pFoo_t pFoo(foo);
        foo->Init(pFoo);
        foo->doSomething();
        foo->doSomethingElse();
    }
    return 0;
}

但我认为有一个更好的解决方案.

Answer 1

不要手动实现.使您的类继承std::enable_shared_from_this并用于std::shared_from_this()获取共享指针.

此外,您应该将对象分配到共享指针:

pFoo_t pFoo = std::make_shared<Foo>();  // good
pFoo_t pFoo(new Foo());                 // legacy

_{(顺便说一句,你要么包括<memory>这个,要么你使用所有这些的TR1版本(然后没有make_shared)并包括<tr1/memory>.我知道MSVC可以让你摆脱很多邋,,但为了便携性你应该选择非此即彼.)}