函数指针类型和值的部分类专门化 [英] Partial class specialization for function pointer type and value

问题描述

我正在使用FLTK来完成与GUI相关的工作，并且它要求将 void（* fn）（Fl_Widget *，void *）类型的函数注册为小部件回调。我厌倦了手工创建函数转发器，将 void * s解压缩为参数并从类中调用适当的静态函数来完成用户请求的工作。

I'm using FLTK to do my GUI related stuff, and it requires functions of type void (*fn)( Fl_Widget*, void* ) to be registered as widget callbacks. I'm tired of creating function forwarders by hand that unpack the void*s into parameters and call appropriate static functions from my classes to do the work the user has requested.

我想出了一个解决方案，但是它需要一个专门针对函数类型和函数地址的类，而这正是我遇到的问题。下面是一些代码：

I came up with a solution, but it requires a class to be specialized for both function type, and function address, and this is where I'm having problems. Here's some code:

template< typename T, typename fn_ptr_t, fn_ptr_t fn_ptr >
struct Fl_Callback_package;

template< typename T, typename Return, typename... Params >
struct Fl_Callback_package< T, Return (*)( Params... ), /* What goes here? */ >
{
  //...
};

编辑：
为了澄清-我不想用特定的功能代替 / *这里是什么？ * / ，但我希望此参数为 Return（* fn_ptr）（Params ...）。

To clarify - I don't want a specific function in place of /* What goes here? */, but rather I would like this parameter to be Return (*fn_ptr)( Params... ). When I try

template< typename T, typename Return, typename... Params >
struct Fl_Callback_package< T, Return (*)( Params... ), Return (*fn_ptr)( Params... ) >

我从GCC 4.8.1收到一个错误，说 fn_ptr 。

I get an error from GCC 4.8.1 saying that fn_ptrwas not declared in this scope.

推荐答案

您将 Return（* fn_ptr）（ Params ...）放在错误的位置。这是部分专业化的模板参数，因此它进入 template< ...> 。

You put Return (*fn_ptr)( Params... ) in the wrong place. It's a template parameter of the partial specialization, so it goes into the template <...>.

template< typename T, typename fn_ptr_t, fn_ptr_t fn_ptr >
struct Fl_Callback_package;

template< typename T, typename Return, typename... Params, Return (*fn_ptr)( Params... ) >
struct Fl_Callback_package< T, Return (*)( Params... ), fn_ptr >
{
  //...
};

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