auto_ptr_ref [英] auto_ptr_ref

问题描述

Hello All，

我的问题是关于自动指针的实现 - 如何

类auto_ptr_ref有效吗？


例如：假设我有：


auto_ptr< int> foo（）;


int bar（）

{

auto_ptr< int *> k = foo（）;

}


我在这里看到的确切调用顺序是（假设没有优化）

-

1）foo返回auto_ptr< int>按值计算

2）在返回值上调用CC。

3）返回的auto_ptr< int>转换为auto_ptr_ref< int>通过

使用运算符auto_ptr_ref

4）现在由于这是一个复制初始化而不是直接的初始化，因此RHS（其类型现在是auto_ptr_ref< int>）需要

转换为auto_ptr< int>再次 - 由auto_ptr< int> c / b的ctor完成，需要auto_ptr_ref< int>

5）现在应该调用CC - 但我们又回到了老问题 -

生成的auto_ptr是临时的，因此不能作为

非const引用传递给函数。


似乎我是在周期的某个地方:-（


Josuttis的书在这方面说不够 -


< quote>

机制依赖于重载

和模板参数扣除规则之间的细微差别。这种差异太微妙了

可用作一般编程工具，但足以启用

auto_ptr类才能正常工作。

< / quote>


有人可以解释一下我没有到达这里吗？

Hello All,
My question is regarding the implementation of auto pointers - How
does the class auto_ptr_ref work?

eg: suppose I have :

auto_ptr<int> foo();

int bar()
{
auto_ptr<int*> k = foo();
}

The exact sequence of calls as I see here is (no optimizations assumed)
-
1) foo returns auto_ptr<int> by value
2) CC is called on the return value.
3) The returned auto_ptr<int> is converted to auto_ptr_ref<int> by
using operator auto_ptr_ref
4) Now since this is a copy initialization and not a direct
initialization, the RHS (whose type is now auto_ptr_ref<int>) needs to
get converted to auto_ptr<int> again - done by ctor of auto_ptr<int>
which takes auto_ptr_ref<int>
5) Now CC should be called - but again we are back to old problem - the
generated auto_ptr is temporary and hence cannot be passed as a
non-const reference to the function.

Seems I am going somewhere in cycle :-(

Josuttis'' book doesnot say enough on this -

<quote>
"The mechanism relies on a slight difference between the overloading
and template argument deduction rules. This difference is too subtle to
be of use as a general programming tool, but it is sufficient to enable
the auto_ptr class to work correctly."
</quote>

Can somebody explain what I am not getting here?

推荐答案

* Neelesh Bodas：

* Neelesh Bodas:

例如：假设我有：
auto_ptr< int> foo（）;

int bar（）
{
auto_ptr< int *> k = foo（）;
}

我在这里看到的确切调用顺序是（假设没有优化）
-
1）foo返回auto_ptr< int> ;按价值


OK。


2）在返回值上调用CC。


不，没有std :: auto_ptr（std :: auto_ptr const&）复制构造函数。


3）返回的auto_ptr< INT>转换为auto_ptr_ref< int>通过
使用operator auto_ptr_ref


OK。


4）现在既然是复制初始化而不是直接
初始化，RHS（其类型现在为auto_ptr_ref< int>）需要转换为auto_ptr< int>再次 - 由auto_ptr< int>
的ctor完成，它采用auto_ptr_ref< int>


好​​的。


5）现在CC应该叫


不，你''已达到目标。


- 但我们又回到了旧问题 -
生成的auto_ptr是临时的，因此无法作为非const传递参考函数。

eg: suppose I have :

auto_ptr<int> foo();

int bar()
{
auto_ptr<int*> k = foo();
}

The exact sequence of calls as I see here is (no optimizations assumed)
-
1) foo returns auto_ptr<int> by value
OK.

2) CC is called on the return value.
No, there is no std::auto_ptr( std::auto_ptr const& ) copy constructor.

3) The returned auto_ptr<int> is converted to auto_ptr_ref<int> by
using operator auto_ptr_ref
OK.

4) Now since this is a copy initialization and not a direct
initialization, the RHS (whose type is now auto_ptr_ref<int>) needs to
get converted to auto_ptr<int> again - done by ctor of auto_ptr<int>
which takes auto_ptr_ref<int>
OK.

5) Now CC should be called
No, you''ve reached the goal.

- but again we are back to old problem - the
generated auto_ptr is temporary and hence cannot be passed as a
non-const reference to the function.

生成auto_ptr的是什么？生成的那个是你命名为''k'的那个。


-

答：因为它弄乱了人们的顺序通常阅读文字。

问：为什么这么糟糕？

A：热门帖子。

问：什么是最多的在usenet和电子邮件中烦人的事情？

What generated auto_ptr? The one generated is the one you''ve named ''k''.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

你好Alf，感谢您的帮助。


实际上，我想的更多在这方面，它正在创造更多的混乱。

Hi Alf, Thanks for the help.

Actually, more I think on this, more confusion it is creating.

auto_ptr< int> foo（）;

int bar（）
{
auto_ptr< int *> k = foo（）;
}

auto_ptr<int> foo();

int bar()
{
auto_ptr<int*> k = foo();
}

当函数返回值时，CC必须得到

调用是不正确的关于返回值（虽然它可能被

编译器优化了？）

不，没有std :: auto_ptr（std :: auto_ptr const&）副本构造函数。

Isn''t is correct that whever a function returns by value, a CC must get
called on the return value(Though it might be optimized away by the
compiler?)

No, there is no std::auto_ptr( std::auto_ptr const& ) copy constructor.

我相信我们不需要std :: auto_ptr（std :: auto_ptr const&）

复制构造函数 - std :: auto_ptr（std :: auto_ptr&）可以完成这项工作 -

，因为foo里面的副本来源是不一定是const：


std :: auto_ptr< int> foo（）

{

std :: auto_ptr< int> v;

返回v; //复制ctor将在v上调用，因为这是返回

值

}


但是当我调试throgh std :: auto_ptr代码我没有找到任何给定拷贝构造函数的调用

（正如你在回复中所说的那样）。


我没有准确到达我错了。

I believe that we do not need a std::auto_ptr( std::auto_ptr const& )
copy constructor - std::auto_ptr( std::auto_ptr& ) would do the work -
since the source of the copy inside "foo" is not necessarily a const :

std::auto_ptr<int> foo()
{
std::auto_ptr<int> v;
return v; // a copy ctor will be called on v since this is return by
value
}

But when I debugged throgh std::auto_ptr code I didnot find any call to
the given copy constructor (as you said in the reply.)

I am not getting where exactly I am going wrong.

* Neelesh Bodas：

* Neelesh Bodas:

嗨Alf，感谢您的帮助。

Hi Alf, Thanks for the help.

Actually, more I think on this, more confusion it is creating.

auto_ptr< int> foo（）;

int bar（）
{
auto_ptr< int *> k = foo（）;
}
一个函数返回值是不正确的，CC必须在返回值上调用（尽管它可能被优化掉了
编译器？）


编号_Some_构造函数必须被调用。该调用可以优化

，但构造函数必须可以调用（as-if规则

进行优化）。


不，没有std :: auto_ptr（std :: auto_ptr const&）复制构造函数。

auto_ptr<int> foo();

int bar()
{
auto_ptr<int*> k = foo();
}
Isn''t is correct that whever a function returns by value, a CC must get
called on the return value(Though it might be optimized away by the
compiler?)
No. _Some_ constructor must be called. That call can be optimized
away, but the constructor must be available to be called (the as-if rule
for optimization).

No, there is no std::auto_ptr( std::auto_ptr const& ) copy constructor.

我相信我们不需要std :: auto_ptr（std） :: auto_ptr const&）
复制构造函数 - std :: auto_ptr（std :: auto_ptr&）可以完成工作 -
因为复制源里面有foo - 不一定是const：

std :: auto_ptr< int> foo（）
{
std :: auto_ptr< int> v;
返回v; //复制ctor将在v上调用，因为这是由
值返回
}

但是当我调试throgh std :: auto_ptr代码时，我没有找到任何对<的调用br />给定的拷贝构造函数（正如你在回复中所说的那样）。

I believe that we do not need a std::auto_ptr( std::auto_ptr const& )
copy constructor - std::auto_ptr( std::auto_ptr& ) would do the work -
since the source of the copy inside "foo" is not necessarily a const :

std::auto_ptr<int> foo()
{
std::auto_ptr<int> v;
return v; // a copy ctor will be called on v since this is return by
value
}

But when I debugged throgh std::auto_ptr code I didnot find any call to
the given copy constructor (as you said in the reply.)

不幸的是，这是一个不同的效果。


这里可以调用引用到非const的复制构造函数，并且我相信_is_在正式级别调用_is_（C ++抽象机器

级别），但是那个调用被你的编译器优化了。


但是，将''v'声明为const，或者使用临时的，你不再是

在可能的复制构造函数调用范围内：_some_构造函数必须是

调用（虽然它可以被优化掉），但是在const或

的情况下临时auto_ptr那个构造函数不是复制构造函数。


我不知道到底哪里出错了。

That is unfortunately a different effect.

Here the reference-to-non-const copy constructor could be called, and I
believe _is_ called at the formal level (the C++ abstract machine
level), but that call is optimized away by your compiler.

However, declare ''v'' as const, or use a temporary, and you''re no longer
in possible copy constructor call territory: _some_ constructor must be
called (although it can be optimized away), but in the case of const or
temporary auto_ptr that constructor isn''t the copy constructor.

I am not getting where exactly I am going wrong.

这次它是将编译器的优化误认为确认

误解了我写的内容（抱歉没有更多

清楚）。我并不是说从不为任何

函数结果调用复制构造函数。我的意思是复制构造函数并不总是要调用

，特别是我的意思，并写道，std :: auto_ptr

没有ref-to-const复制构造函数，这是用于函数结果的通常复制

构造函数。


-

答：因为它弄乱了人们通常阅读文字的顺序。

问：为什么这么糟糕？

A：热门发布。

问：usenet和电子邮件中最烦人的事情是什么？

This time it was mistaking the compiler''s optimization as confirmation
of a misinterpretation of what I wrote (sorry about not being more
clear). I didn''t mean that a copy constructor is never called for any
function result. I meant that a copy constructor does not always have
to be called, and in particular I meant, and wrote, that std::auto_ptr
doesn''t have a ref-to-const copy constructor, which is the usual copy
constructor used for a function result.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

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