C++ 如何在命名空间 std 中转发声明模板类？
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How to forward declare a template class in namespace std?
提问by nakiya
#ifndef __TEST__
#define __TEST__
namespace std
{
template<typename T>
class list;
}
template<typename T>
void Pop(std::list<T> * l)
{
while(!l->empty())
l->pop();
}
#endif
and used that function in my main. I get errors. Of course, I know that there are more template params for std::list(allocator I think). But, that is beside the point. Do I have to know the full template declaration of a template class to be able to forward declare it?
并在我的主要功能中使用了该功能。我收到错误。当然，我知道有更多的模板参数std::list（我认为是分配器）。但是，这不是重点。我是否必须知道模板类的完整模板声明才能转发声明它？
EDIT: I wasn't using a pointer before - it was a reference. I'll try it out with the pointer.
编辑：我之前没有使用指针 - 它是一个引用。我会用指针试一试。
回答by Jon Purdy
The problem is not that you can't forward-declare a template class. Yes, you do need to know all of the template parameters and their defaultsto be able to forward-declare it correctly:
问题不在于您不能向前声明模板类。是的，您确实需要知道所有模板参数及其默认值才能正确地向前声明它：
namespace std {
template<class T, class Allocator = std::allocator<T>>
class list;
}
But to make even such a forward declaration in namespace stdis explicitly prohibited by the standard: the onlything you're allowed to put in stdis a template specialisation, commonly std::lesson a user-defined type. Someone else can cite the relevant text if necessary.
但是，为了使即使是这样在向前声明namespace std由该标准明确禁止：在只有你允许把在事情std是一个模板专业化，通常std::less在用户定义类型。如有必要，其他人可以引用相关文本。
Just #include <list>and don't worry about it.
只是#include <list>不要担心。
Oh, incidentally, any name containing double-underscores is reserved for use by the implementation, so you should use something like TEST_Hinstead of __TEST__. It's not going to generate a warning or an error, but if your program has a clash with an implementation-defined identifier, then it's not guaranteed to compile or run correctly: it's ill-formed. Also prohibited are names beginning with an underscore followed by a capital letter, among others. In general, don't start things with underscores unless you know what magic you're dealing with.
哦，顺便说一句，任何包含双下划线的名称都保留供实现使用，因此您应该使用类似的名称TEST_H而不是__TEST__. 它不会生成警告或错误，但如果您的程序与实现定义的标识符发生冲突，则不能保证它可以正确编译或运行：它是格式错误的。还禁止以下划线开头的名称，后跟大写字母等。通常，除非您知道要处理的魔术，否则不要以下划线开头。
回答by user1638075
I solved that problem.
我解决了那个问题。
I was implementing an OSI Layer (slider window, Level 2) for a network simulation in C++ (Eclipse Juno). I had frames (template <class T>) and its states (state pattern, forward declaration).
我正在为 C++ (Eclipse Juno) 中的网络模拟实现 OSI 层（滑块窗口，级别 2）。我有框架（模板<class T>）及其状态（状态模式，前向声明）。
The solution is as follows:
解决方法如下：
In the *.cppfile, you must include the Header file that you forward, i.e.
在*.cpp文件中必须包含你转发的Header文件，即
ifndef STATE_H_
#define STATE_H_
#include <stdlib.h>
#include "Frame.h"
template <class T>
class LinkFrame;
using namespace std;
template <class T>
class State {
protected:
LinkFrame<int> *myFrame;
}
Its cpp:
它的 cpp：
#include "State.h"
#include "Frame.h"
#include "LinkFrame.h"
template <class T>
bool State<T>::replace(Frame<T> *f){
And... another class.
还有……另一堂课。
回答by Grozz
Forward declaration should have complete template arguments list specified.
前向声明应指定完整的模板参数列表。
回答by Arne
there is a limited alternative you can use
您可以使用的替代方案有限
header:
标题：
class std_int_vector;
class A{
std_int_vector* vector;
public:
A();
virtual ~A();
};
cpp:
cp：
#include "header.h"
#include <vector>
class std_int_vector: public std::vectror<int> {}
A::A() : vector(new std_int_vector()) {}
[...]
not tested in real programs, so expect it to be non-perfect.
没有在实际程序中测试过，所以期望它是不完美的。
