对非const的引用，以绑定到临时对象 [英] A reference to non-const to be bound to a temporary object
问题描述
您好：
我不确定如何正确地说出这个问题，所以我先从
开始列出我的代码遇到问题：
int main（）
{
std :: vector< int> x;
x.swap（std :: vector< int>（））;
返回0;
}
这也是不允许的：
std :: vector< int>得到（）
{
返回std :: vector< int>（）;
}
int main（）
{
std :: vector< int> x;
x.swap（get（））;
返回0;
}
我被告知在另一个新闻组，以下代码是
非标准C ++。原因是标准C ++
不允许将非const引用绑定到
临时对象。
但是能够将对非const的引用绑定到临时的
对象对于利用交换方法很有用。
我发现能够避免这种限制是有用的，因为它允许我包装昂贵的资源来创建
并销毁到一个类然后移动
使用交换功能最大限度地减少复制和保证异常安全
所有这些都是简洁的代码。
例如，如果Resource类包含一个昂贵的
资源，并且create static方法创建资源
，而open static方法打开资源：
//资源接口：
资源资源:: open（char * name）;
资源资源:: create（char * name）;
//部分代码：
资源1;
资源2;
resource1.swap （Resource :: open（" XYZ"））;
resource2.swap（Resource :: create（" abc"））;
I因此，我对这种限制的动机感兴趣，并且如果还有另一种方法可以同样有效地完成同样的工作，那么就会感兴趣。
-John
Hello:
I''m not sure how to word this question properly, so I''ll start by
listing the code I am having problems with:
int main()
{
std::vector<int> x;
x.swap(std::vector<int>());
return 0;
}
This is also not allowed:
std::vector<int> get()
{
return std::vector<int>();
}
int main()
{
std::vector<int> x;
x.swap(get());
return 0;
}
I was told on another newsgroup that the following code is
nonstandard C++. The reason being that "Standard C++
doesn''t allow a reference to non-const to be bound to a
temporary object."
But being able to bind a reference to non-const to a temporary
object is useful for taking advantage of using a swap method.
I find it useful to be able to avoid this restriction because
it allows me to wrap resources that are expensive to create
and destroy into a class and then move it around while
minimising copying and guaranteeing exception safety by
using a swap function, all in a concise piece of code.
For instance if the Resource class wrapped an expensive
resource and the create static method creates the resource
while the open static method opens the resource:
// Resource interface:
Resource Resource::open(char *name);
Resource Resource::create(char *name);
// Some code:
Resource resource1;
Resource resource2;
resource1.swap(Resource::open("XYZ"));
resource2.swap(Resource::create("abc"));
I''m therefore interested in the motivation for this restriction and
if there is another way to do the same job just as efficiently.
-John
推荐答案
John Ky在新闻中写道：10 *************** @ cousin.sw.oz.au：
John Ky wrote in news:10***************@cousin.sw.oz.au:
你好：
我不确定如何正确地说出这个问题，所以我先从
列出代码我遇到问题：
int main（）
{st / :: std :: vector< int> x;
x.swap（std :: vector< int>（））;
更改为：
std :: vector< ont>（）。swap（x）;
返回0;
}
Hello:
I''m not sure how to word this question properly, so I''ll start by
listing the code I am having problems with:
int main()
{
std::vector<int> x;
x.swap(std::vector<int>());
change to:
std::vector<ont>().swap( x );
return 0;
}
HTH。
Rob。
-
http：//www.victim-prime.dsl.pipex .com /
John Ky写道：
John Ky wrote:
...
我因此对动机感兴趣这个限制和
如果还有另一种方法可以同样有效地完成同样的工作。
...
...
I''m therefore interested in the motivation for this restriction and
if there is another way to do the same job just as efficiently.
...
只是以其他方式做到这一点
std :: vector< int> x;
std :: vector< int>（）。swap（x）;
或在您的情况下
资源1;
资源2;
资源::打开（" XYZ"）。swap（resource1）;
资源::创建（" abc"）。swap（resource2）;
-
祝你好运，
Andrey Tarasevich
Just do it other way around
std::vector<int> x;
std::vector<int>().swap(x);
or in your case
Resource resource1;
Resource resource2;
Resource::open("XYZ").swap(resource1);
Resource::create("abc").swap(resource2);
--
Best regards,
Andrey Tarasevich
John Ky写道：
< snip> intro< />
John Ky wrote:
<snip> intro </>
int main（）
{st / :: std :: vector< int> x;
x.swap（std :: vector< int>（））;
返回0;
}
< snip>更多示例代码< />
我在另一个新闻组中被告知以下代码是非标准C ++。原因是标准C ++
不允许将非const引用绑定到
临时对象。
但是能够绑定对非临时
对象的非const引用对于利用交换方法很有用。
< snip />
因此，我对此限制的动机感兴趣，如果还有其他办法可以做同样的工作同样有效。
int main()
{
std::vector<int> x;
x.swap(std::vector<int>());
return 0;
}
<snip> more example code </>
I was told on another newsgroup that the following code is
nonstandard C++. The reason being that "Standard C++
doesn''t allow a reference to non-const to be bound to a
temporary object."
But being able to bind a reference to non-const to a temporary
object is useful for taking advantage of using a swap method.
<snip/>
I''m therefore interested in the motivation for this restriction and
if there is another way to do the same job just as efficiently.
在这种特殊情况下，x.clear（）可能是最有效的和简单的b $ b简洁选项。
In this particular case, x.clear( ) is probably the most efficient and
concise option.
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