如何从c代码调用C ++函数？ [英] How to call a C++ function from c code?
问题描述
如何从c代码调用C ++函数？
externC ++？
我认为它只是禁止警告。
我可能要考虑c ++对象创建问题。
C ++是否需要是一个静态函数？
BTW，c代码用C ++编译器编译。
How to call a C++ function from c code?
extern "C++"?
I think it just inhibit the warning.
I maybe have to consider c++ object creation problem.
Does the C++ need to be a static function?
BTW, the c code is compiled with C++ compiler.
推荐答案
" AL @ TW"写道：
"AL@TW" wrote:
>
如何从c代码调用C ++函数？
extern" C ++" ;?
>
How to call a C++ function from c code?
extern "C++"?
No.一般来说，你不能，因为C ++重载。你可以用$ C $ b从C ++调用C.然后。
-
[邮件]：Chuck F（cinefalconer at maineline dot net）
[page]：< http://cbfalconer.home.att.net>
尝试下载部分。
No. In general, you can''t, because of C++ overloading. You can
call C from C++, however.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
AL @ TW写道：
AL@TW wrote:
如何从c代码调用C ++函数？
extern C ++？
我认为它只是禁止警告。
我可能要考虑c ++对象创建问题。
How to call a C++ function from c code?
extern "C++"?
I think it just inhibit the warning.
I maybe have to consider c++ object creation problem.
这是对comp.lang.c的OT; comp.lang.c ++在大厅的左边。
< OT>
我认为你会得到的答案是C ++函数声明
应该包含在externC中。 {...}阻塞在C ++中编译时，
而C编译器看到的原型声明应该_not_
如此封闭。通常的方法是在单个
标头中使用BEGIN_C_DECLS和
END_C_DECLS宏（Google用于惯用定义），这两种语言的编译器都可以使用它。
我不记得C ++编译器是否需要使用externC封装的函数定义
{...}或是否会记住只需通过附上声明即可编译函数C风格。
This is OT for comp.lang.c; comp.lang.c++ is down the hall, to the left.
<OT>
The answer I assume you''ll get is that the C++ function declaration
should be enclosed in an extern "C" { ... } block when compiled in C++,
while the prototype declaration that the C compiler sees should _not_ be
so enclosed. The usual way to do this is with the BEGIN_C_DECLS and
END_C_DECLS macros (Google for the idiomatic definitions) in a single
header that can be used by compilers of both languages.
I don''t recall if the C++ compiler needs the function definition to be
enclosed with extern "C" { ... } or whether it will "remember" to
compile the function C-style simply by having the declaration enclosed.
C ++是否需要是静态函数？
Does the C++ need to be a static function?
否但，C ++的某些功能在一个函数中是不可用的，因为
被编译为externC。 ，就像默认参数一样，变量
签名等。我很确定没有办法用类
方法来做，无论是静态的还是不，只是简单的功能。常见的
策略是具有C兼容的包装器。函数，它反过来调用真正的C ++函数。 （例如，一个对象可以通过void *参数传递给
包装器，然后包装器将它转换为
对象并调用一个方法结果。）
< / OT>
No. However, some features of C++ aren''t available in a function that
is being compiled as extern "C", like default arguments, variable
signatures, etc. I''m pretty sure there''s no way to do it with a class
method either, whether static or not, just plain functions. A common
tactic is to have a C-compatible "wrapper" function, which in turn calls
the real C++ functions. (For instance, an object may be passed in to
the wrapper via a void* parameter, and the wrapper then casts it to an
object and invokes a method on the result.)
</OT>
BTW，c代码是用C ++编译器编译的。
BTW, the c code is compiled with C++ compiler.
如果你用C ++编译器编译它，代码就是C ++。你可能
已经使用了与C相同的C ++子集（并且希望
具有相同的含义，但这并不保证），但它不是'除非你使用C编译器获得
，否则将获得。
S
If you''re compiling it with a C++ compiler, the code is C++. You may
have used a subset of C++ that looks the same as C (and hopefully has
the same meaning, which isn''t guaranteed), but it isn''t C unless you''re
using a C compiler.
S
" ; AL @ TW" < schosnab ... @ gmail.comwrote：
"AL@TW" <schosnab...@gmail.comwrote:
如何从c代码调用C ++函数？
How to call a C++ function from c code?
你不是。 [你可以在某些情况下，但细节
是特定于实现的，并且超出了C语言本身的范围，因此在comp.lang.c中是偏离主题的。 ]
You don''t. [You can in certain cases, but the details
are implementation specific and outside the scope of
the C language itself, hence off-topic in comp.lang.c.]
extern" C ++"？
我认为它只是禁止警告。
我可能不得不考虑c ++ *对象创建问题。
C ++是否需要是一个静态函数？
extern "C++"?
I think it just inhibit the warning.
I maybe have to consider c++ *object creation problem.
Does the C++ need to be a static function?
在comp.lang.c ++中询问C ++。
Ask about C++ in comp.lang.c++.
BTW，c代码是用C ++编译器。
BTW, the c code is compiled with C++ compiler.
然后你没有C语言问题，而是C ++
问题。再次，请在comp.lang.c ++中询问。
-
Peter
Then you don''t have a C language question, but a C++
question. Again, ask in comp.lang.c++.
--
Peter
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