c 11 – 为什么C标准不禁止这种可怕的使用？

发布时间：2020-12-16 09:31:06 所属栏目：百科来源：网络整理

导读：源代码非常简单且不言而喻.问题包含在评论中. #include iostream#include functionalusing namespace std;using namespace std::tr1;struct A{ A() { cout "A::ctor" endl; } ~A() { cout "A::dtor" endl; } void foo() {}};int main(){ A a; /* Performance

源代码非常简单且不言而喻.问题包含在评论中.

#include <iostream>
#include <functional>

using namespace std;
using namespace std::tr1;

struct A
{
    A()
    {
        cout << "A::ctor" << endl;
    }

    ~A()
    {
        cout << "A::dtor" << endl;
    }

    void foo()
    {}
};

int main()
{
    A a;
    /*
    Performance penalty!!!

    The following line will implicitly call A::dtor SIX times!!! (VC++ 2010)    
    */
    bind(&A::foo,a)();  

    /*
    The following line doesn't call A::dtor.

    It is obvious that: when binding a member function,passing a pointer as its first 
    argument is (almost) always the best way. 

    Now,the problem is: 

    Why does the C++ standard not prohibit bind(&SomeClass::SomeMemberFunc,arg1,...) 
    from taking arg1 by value? If so,the above bind(&A::foo,a)(); wouldn't be
    compiled,which is just we want.
    */
    bind(&A::foo,&a)(); 

    return 0;
}

解决方法

首先,您的代码有第三种替代方案：

bind(&A::foo,std::ref(a))();

现在,为什么默认情况下复制采用参数？我认为,但这只是一个疯狂的猜测,认为绑定默认行为与参数生存期无关是优选的：绑定的结果是一个仿函数,其参数销毁后可能会延迟调用.

您是否希望以下代码默认产生UB？

void foo(int i) { /* ... */ }

int main()
{
    std::function<void ()> f;

    {
        int i = 0;
        f = std::bind(foo,i);
    }

    f(); // Boom ?
}

（编辑：李大同）

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