c – boost :: bind内部副本/副本?

我希望了解boost :: bind做什么样的函数对象的内部副本.由于这些对象的构造函数似乎没有被调用,我认为这是一种“非常浅的副本”,所以我引入了动态内存分配来产生一些错误.但是,下面代码的运行时输出似乎表示对bind生成的内部副本有三个额外的析构函数调用.

using namespace std;
using namespace boost;

class M{
    int *somedata;
public:
    M(){ somedata=new int[5]; cout<<"from cstr\n"; somedata[1]=0;}
    ~M(){cout<<"from dstr\n"; delete somedata;}

    int operator()(int i){ cout<<++somedata[i]<<endl; return 0;}
};


int main()
{
    M instM;

    bind<int>(instM,1)();
    //bind<int>(&M::operator(),&instM,1)(); //this works with no errors, of course

    instM(1); //would not change the order of output

    return 0;
}

输出……提出了一些额外的难题 – 例如.为什么在调用operator()之前第一个dstr事件发生?在最后一次失败的析构函数调用之前还要注意“2”.

from cstr
from dstr
1
from dstr
bind_copy(73365) malloc: *** error for object 0x1001b0: double free
*** set a breakpoint in malloc_error_break to debug
from dstr
bind_copy(73365) malloc: *** error for object 0x1001b0: double free
*** set a breakpoint in malloc_error_break to debug
2
from dstr
bind_copy(73365) malloc: *** error for object 0x1001b0: double free
*** set a breakpoint in malloc_error_break to debug

所以问题是:任何人都可以简单地解释一下这个顺序,以及哪种副本可以绑定?

…经过一番思考后,我意识到bind只是使用(这里是默认的)复制构造函数.在提供了这个cstr的一些自定义版本(带有内存分配和as-deep-as-one-wish许可版本的副本)后,输出变得干净(应该如此),但是谜题仍然存在:复制构造函数有三个调用.所以在这种情况下,boost :: bind会生成函数对象的三个副本.为什么以及以何种顺序? (对于嵌套的boost :: binds,这可能导致内部副本数量急剧增长.)

定义了cp-cstr的输出,并添加了一些“遗产标记”(“P”=父,每个cp cstr添加“-C”):

 from cstr P
 from cp cstr P-C
 from cp cstr P-C-C
 from cp cstr P-C-C-C
 from dstr P-C-C
 P-C-C-C:1
 from dstr P-C-C-C
 from dstr P-C
 P:1
 from dstr P

最佳答案 见
here

By default, bind makes a copy of the provided function object. boost::ref and boost::cref can be used to make it store a reference to the function object, rather than a copy. This can be useful when the function object is noncopyable, expensive to copy, or contains state; of course, in this case the programmer is expected to ensure that the function object is not destroyed while it’s still being used.

点赞