对于Windows HANDLE，使用std :: unique_ptr [英] Using std::unique_ptr for Windows HANDLEs

问题描述

我尝试使用std :: unique_ptrs以异常安全的方式管理Windows HANDLE。

首先我尝试：

  struct HandleDeleter 
 {
 void operator（）（HANDLE handle）
 {
 if（handle）
 {
 FindVolumeClose（handle）
} 
} 
} 
 typedef std :: unique_ptr< HANDLE，HandleDeleter> unique_vol_handle_t; 
  

后来在我的代码中，当我尝试使用它：

unique_vol_handle_t volH（FindFirstVolumeW（buffer，MAX_GUID_PATH））;

Visual Studio 2012RC：

  1>错误C2664：'std :: unique_ptr <_Ty，_Dx> :: unique_ptr（std :: nullptr_t）throw（）'：无法将参数1从'HANDLE'转换为'std :: nullptr_t'
 1& with 
 1> [
 1> _Ty = HANDLE，
 1> _Dx = VolumeHandleDeleter 
 1> ] 
 1> nullptr只能转换为引用上述volH声明行的指针或句柄类型
  

搜索一段时间后，我发现博客文章基本上说，添加：

typedef HANDLE pointer; / code>

到结构声明的顶部，一切都会很好。

我不相信，但我试过它，它确实解决了错误。我很困惑如何定义一个类型（没有引用它）可能会产生这样的区别。

两个问题：

1）您能解释原始错误吗？我不明白为什么编译器引用 std :: nullptr_t / nullptr 。

2） typedef解决这个（或至少看起来）？

解决方案

unique_ptr的实现检查删除器上是否存在 :: pointer 类型。如果删除器有一个 :: pointer 类型，那么这个类型被用作指针 typedef在 unique_ptr 。

根据cppreference.com， unique_ptr ::指针类型定义为

std :: remove_reference< D> :: type :: pointer 如果该类型存在，否则 T *

I am attempting to use std::unique_ptrs to manage Windows HANDLEs in an exception-safe manner.

First I tried:

struct HandleDeleter
{
    void operator()( HANDLE handle )
    {
        if( handle )
        {
            FindVolumeClose( handle )
        }
    }
}
typedef std::unique_ptr< HANDLE, HandleDeleter > unique_vol_handle_t;

Later in my code when I try to use it:

unique_vol_handle_t volH( FindFirstVolumeW( buffer, MAX_GUID_PATH ) );

I get the following error from Visual Studio 2012RC:

1>          error C2664: 'std::unique_ptr<_Ty,_Dx>::unique_ptr(std::nullptr_t) throw()' : cannot convert parameter 1 from 'HANDLE' to 'std::nullptr_t'
1>          with
1>          [
1>              _Ty=HANDLE,
1>              _Dx=VolumeHandleDeleter
1>          ]
1>          nullptr can only be converted to pointer or handle types

referencing the volH declaration line, immediately above.

After searching for some time, I found a blog article which basically says, add:

typedef HANDLE pointer;

to the top of the struct declaration, and all will be well.

I didn't believe it, but I tried it and it did resolve the error. I'm puzzled how defining a type (without even referencing it) could make such a difference.

Two questions:

1) Can you explain the original error? I don't understand why the compiler is referring to std::nullptr_t/nullptr.

2) How is it that the typedef resolves this (or at least appears to)? Is there a less 'spooky action at a distance' solution to this?

解决方案

The implementation of unique_ptr checks for the presence of a ::pointer type on the deleter. If the deleter has a ::pointer type then this type is used as the pointer typedef on the unique_ptr. Otherwise a pointer to the first template argument is used.

According to cppreference.com, the unique_ptr::pointer type is defined as

std::remove_reference<D>::type::pointer if that type exists, otherwise T*

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