运算符在模板中重载 [英] operator overloading in templates
问题描述
我在模板中添加了一个操作符overloader
class：
graph.h ：
模板< class NODE>
class Node
{
朋友ostream&运营商LT;< （ostream&，Node< NODE>&）; // 80行
NODE * info;
[snip]
}
图。 cpp：
[snip]
模板< class NODE>
ostream& operator<<（ostream& output，const Node< ; NODE>&节点）
{
输出<< *（node.info）<< " " ;;
返回输出;
}
[snip}
使用gcc编译时我收到警告：
来自graph.cpp的文件：2：
... / graph.h：80：警告：朋友声明`std :: ostream&
operator<<（std :: ostream&，Node< NODE>&）''声明一个
非模板函数
... /graph.h:80：警告:(如果这不是您的意图，请确保已经声明
功能模板并在
$ b $之后添加<> b函数名称在这里）-Wno-non-template-friend禁用此警告
警告的原因是什么，我该如何避免？
谢谢。
问候，
克里斯
Christian Christmann写道：我在模板中添加了一个操作符overloader
类：
graph.h：
在这里添加：
tem板< class NODE> class Node;
template< class NODE> ostream的和放;运算符<（&ostream&，Node< NODE>&）;
并且，运营商<<<<<<<<<<<<<<<<<需要它第二个
参数作为对* non-const * Node<>？
模板的引用< class NODE>
类Node
{
朋友ostream& operator<< （ostream&，Node< NODE>&）; //第80行
NODE * info;
[snip]
}
graph.cpp：
[snip]
template< class NODE>
ostream& operator<<（ostream& output，const Node< NODE>& node）
{
output<< *（node.info）<< " " ;;
返回输出;
}
[snip}
使用gcc编译时我收到警告：
在graph.cpp中包含的文件中：2：
../graph.h：80：警告：朋友声明`std :: ostream&
运算符<<（std :: ostream&，Node< NODE>&）''声明一个
非模板函数
../graph.h：80：警告:(如果这不是你想要的，请确保
函数模板已经声明并添加< ;>此后
函数名称）-Wno-non-template-friend禁用此警告
该警告的原因是什么？如何避免？
声明您的运营商<<作为类模板之前的模板。见上面的
。
V
" Victor Bazarov" <五******** @ comAcast.net>在消息中写道
news：s％******************* @ newsread1.mlpsca01.us.t o.verio.net ...Christian Christmann写道：我在模板中添加了一个操作符overloader
类：
graph.h：
在这里添加：
模板< class NODE> class Node;
template< class NODE> ostream的和放; operator<（&ostream&，Node< NODE>&）;
为什么这些声明是必要的？我注意到VS 2003和VC ++都不需要它们。
template< class NODE>
class Node
朋友ostream& operator<< （ostream&，Node< NODE>&）; //第80行
NODE * info;
[snip]
}
graph.cpp：
[snip]
template< class NODE>
ostream& operator<<（ostream& output，const Node< NODE>& node）
{
output<< *（node.info）<< " " ;;
返回输出;
}
[snip}
DW
David White写道：" Victor Bazarov" <五******** @ comAcast.net>在消息中写道
新闻：s％******************* @ newsread1.mlpsca01.us.t o.verio.net ...
Christian Christmann写道：
我在模板中添加了一个操作符overloader
类：
graph.h：
在此处添加：
模板< class NODE> class Node;
template< class NODE> ostream的和放; operator<（&ostream&，Node< NODE>&）;
为什么这些声明是必要的？我注意到VS 2003和VC ++都不需要它们。
标准，在14.5.3 / 1中，如果朋友函数声明
不是模板声明（不是以''模板''
关键字开头），而且该函数的名称是模板ID，朋友
声明是指模板特化，如果它不是模板ID
而是''unqualified-id''，它声明了一个普通的函数。所以，如果你（或者OP
）不要声明运营商<<事先作为模板，它被认为是一个简单的函数，这违背了它的第二个
参数依赖于''NODE''这一事实。
如果''运算符<<''是模板ID，则解决冲突，通过声明''运算符<<'''模板来实现
这是''Node''上面的第二个模板
声明。声明''Node''作为类模板
（第一个声明语句）是必要的，因为它在
中用于以下函数模板声明。
template< class NODE>
class Node
朋友ostream& operator< ;< （ostream&，Node< NODE>&）; //第80行
NODE * info;
[snip]
}
graph.cpp：
[snip]
template< class NODE>
ostream& operator<<（ostream& output，const Node< NODE>& node）
{
output<< *（node.info）<< " " ;;
返回输出;
}
[snip}
DW
V
Hi,
I added an operator overloader in my template
class:
graph.h:
template <class NODE>
class Node
{
friend ostream &operator<< (ostream &, Node<NODE> &); //line 80
NODE* info;
[snip]
}
graph.cpp:
[snip]
template <class NODE>
ostream &operator<<(ostream &output, const Node<NODE> &node)
{
output << *(node.info) << " ";
return ouput;
}
[snip}
While compiling with gcc I get the warning:
In file included from graph.cpp:2:
.../graph.h:80: warning: friend declaration `std::ostream&
operator<<(std::ostream&, Node<NODE>&)'' declares a
non-template function
.../graph.h:80: warning: (if this is not what you intended, make sure the
function template has already been declared and add <> after the
function name here) -Wno-non-template-friend disables this warning
What is the reason for that warning and how can I avoid it?
Thank you.
Regards,
Chris
Christian Christmann wrote:I added an operator overloader in my template
class:
graph.h:
Add here:
template<class NODE> class Node;
template<class NODE> ostream& operator<<(ostream&, Node<NODE>&);
And, is there any particular reason why your operator<< needs it second
argument as a reference to a *non-const* Node<>?
template <class NODE>
class Node
{
friend ostream &operator<< (ostream &, Node<NODE> &); //line 80
NODE* info;
[snip]
}
graph.cpp:
[snip]
template <class NODE>
ostream &operator<<(ostream &output, const Node<NODE> &node)
{
output << *(node.info) << " ";
return ouput;
}
[snip}
While compiling with gcc I get the warning:
In file included from graph.cpp:2:
../graph.h:80: warning: friend declaration `std::ostream&
operator<<(std::ostream&, Node<NODE>&)'' declares a
non-template function
../graph.h:80: warning: (if this is not what you intended, make sure the
function template has already been declared and add <> after the
function name here) -Wno-non-template-friend disables this warning
What is the reason for that warning and how can I avoid it?
Declare your operator<< as a template before the class template. See
above.
V
"Victor Bazarov" <v.********@comAcast.net> wrote in message
news:s%*******************@newsread1.mlpsca01.us.t o.verio.net...Christian Christmann wrote:I added an operator overloader in my template
class:
graph.h:
Add here:
template<class NODE> class Node;
template<class NODE> ostream& operator<<(ostream&, Node<NODE>&);
Why are these declarations necessary? I notice that neither VS 2003 nor VC++
6.0 needs them.
template <class NODE>
class Node
{
friend ostream &operator<< (ostream &, Node<NODE> &); //line 80
NODE* info;
[snip]
}
graph.cpp:
[snip]
template <class NODE>
ostream &operator<<(ostream &output, const Node<NODE> &node)
{
output << *(node.info) << " ";
return ouput;
}
[snip}
DW
David White wrote:"Victor Bazarov" <v.********@comAcast.net> wrote in message
news:s%*******************@newsread1.mlpsca01.us.t o.verio.net...Christian Christmann wrote:I added an operator overloader in my template
class:
graph.h:
Add here:
template<class NODE> class Node;
template<class NODE> ostream& operator<<(ostream&, Node<NODE>&);
Why are these declarations necessary? I notice that neither VS 2003 nor VC++
6.0 needs them.
The Standard, in 14.5.3/1, says that if the friend function declaration
that is not a template declaration (does not begin with the ''template''
keyword) and the name of the function is a template-id, the friend
declaration refers to a template specialisation, if it''s not a template-id
but an ''unqualified-id'', it declares an ordinary function. So, if you (or
the OP) don''t declare operator<< as a template beforehand, it is
considered a simple function, which goes against the fact that its second
argument depends on ''NODE''.
The conflict is resolved if ''operator<<'' is a template-id, which is
achieved by declaring ''operator<<'' a template (that''s the second template
declaration above ''Node''). The declaration of ''Node'' as a class template
(the very first declaration statement) is necessary because it''s used in
the following function template declaration.
template <class NODE>
class Node
{
friend ostream &operator<< (ostream &, Node<NODE> &); //line 80
NODE* info;
[snip]
}
graph.cpp:
[snip]
template <class NODE>
ostream &operator<<(ostream &output, const Node<NODE> &node)
{
output << *(node.info) << " ";
return ouput;
}
[snip}
DW
V
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