大括号在 Boost::extension 中如何工作，如何自己制作这样的宏？

发布于 2024-11-28 05:57:34 字数 790 浏览 1 评论 0原文

我看看我们如何使用 Boost::Extension BOOST_EXTENSION_TYPE_MAP_FUNCTION 宏。

例如 this：

BOOST_EXTENSION_TYPE_MAP_FUNCTION
{
    std::map<std::string, boost::extensions::factory<service> > &factories(types.get());
    factories["file_service"].set<file_service>();
}

BOOST_EXTENSION_TYPE_MAP_FUNCTION 宏在扩展名.hpp。

我想知道这个宏如何理解大括号中的内容，以及如何将这个宏扩展为诸如“Hello 扩展宏”之类的内容？

原文

I look at how we use Boost::Extension BOOST_EXTENSION_TYPE_MAP_FUNCTION macro.

For example like this:

BOOST_EXTENSION_TYPE_MAP_FUNCTION
{
    std::map<std::string, boost::extensions::factory<service> > &factories(types.get());
    factories["file_service"].set<file_service>();
}

BOOST_EXTENSION_TYPE_MAP_FUNCTION macro is defined in extension.hpp.

I wonder how this macro understands what is in Curly Brackets and how for example expand this macro to something that would cout anything like "Hello extended macro"?

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倚栏听风 2024-12-05 05:57:34

让我将我的评论放入答案中......

宏是对编译器的指令（我在这里使用集体术语）在该位置替换定义为该宏的符号，例如

#define FOO 1

int val = FOO; // at this point, FOO is replaced with 1

（ps请不要在C++）

现在，您的情况发生的是有一组符号（函数的签名）定义为宏，因此所发生的只是编译器将用符号替换宏，最终结果是看起来（大致）像这样：

void boost_extension_exported_type_map_function(boost::extensions::type_map& types)
{
    std::map<std::string, boost::extensions::factory<service> > &factories(types.get());
    factories["file_service"].set<file_service>();
}

正如你所看到的，这是一个简单的功能。你也可以这样做（但除非你有一个很好的理由，否则不要这样做）

#define BOB void foo(std::string const& bar)

BOB
{
  std::cout << "HEllo: " << bar << std::endl;
}

它只是允许用户为该函数定义自己的实现......大概在其他地方 - 它获取该函数的地址并通过指针...

Let me put my comment into an answer...

A macro is an instruction to the compiler (I use the collective term here) to substitute at that location the symbols defined as that macro, for example

#define FOO 1

int val = FOO; // at this point, FOO is replaced with 1

(p.s. please don't do this in C++)

Now, what is happening in your case is that there is a set of symbols (the signature of a function) defined as a macro, so all that happens is the compiler will substitute the macro with the symbols, and the end result would look (roughly) like this:

void boost_extension_exported_type_map_function(boost::extensions::type_map& types)
{
    std::map<std::string, boost::extensions::factory<service> > &factories(types.get());
    factories["file_service"].set<file_service>();
}

Which as you can see is a simple function. You can do this too (but don't unless you have a very good reason)

#define BOB void foo(std::string const& bar)

BOB
{
  std::cout << "HEllo: " << bar << std::endl;
}

It simply allows a user to define their own implementation for that function... presumably somewhere else - it takes the address of that function and uses it via a pointer...

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