出于这个答案的目的,我在 gcc 上搜索了一个非常旧的版本,因为它更小,更容易推理。在最新版本或其他编译器中的实现可能完全不同。
搜索到的源文件可以在opensource.apple.com查看
gcc 有一个函数is_friend(),定义在 friend.c 中,如果函数或类型是朋友,它基本上返回。下面是函数的相关代码:
int
is_friend (type, supplicant)
tree type, supplicant;
{
int declp;
register tree list;
tree context;
if (supplicant == NULL_TREE || type == NULL_TREE)
return 0;
/*
Comment added by me: The following defines are in tree.h
#define TREE_CODE(NODE) ((enum tree_code) (NODE)->common.code)
#define TREE_CODE_CLASS(CODE) tree_code_type[(int) (CODE)]
This is expanded as:
declp = (tree_code_type[(int) (((enum tree_code) (supplicant)->common.code))] == 'd')
*/
declp = (TREE_CODE_CLASS (TREE_CODE (supplicant)) == 'd');
// That is, it will simply search the array for the code of the supplicant and check if it is a function declaration.
if (declp)
/* It's a function decl. */
{
tree list = DECL_FRIENDLIST (TYPE_MAIN_DECL (type));
tree name = DECL_NAME (supplicant);
tree ctype;
if (DECL_FUNCTION_MEMBER_P (supplicant))
ctype = DECL_CLASS_CONTEXT (supplicant);
else
ctype = NULL_TREE;
for (; list ; list = TREE_CHAIN (list))
{
if (name == FRIEND_NAME (list))
{
tree friends = FRIEND_DECLS (list);
for (; friends ; friends = TREE_CHAIN (friends))
{
if (same_type_p (ctype, TREE_PURPOSE (friends)))
return 1;
if (TREE_VALUE (friends) == NULL_TREE)
continue;
if (supplicant == TREE_VALUE (friends))
return 1;
/* With -fguiding-decls we are more lenient about
friendship. This is bogus in general since two
specializations of a template with non-type
template parameters may have the same type, but
be different.
Temporarily, we are also more lenient to deal
with nested friend functions, for which there can
be more than one FUNCTION_DECL, despite being the
same function. When that's fixed, the
FUNCTION_MEMBER_P bit can go. */
if ((flag_guiding_decls
|| DECL_FUNCTION_MEMBER_P (supplicant))
&& same_type_p (TREE_TYPE (supplicant),
TREE_TYPE (TREE_VALUE (friends))))
return 1;
if (TREE_CODE (TREE_VALUE (friends)) == TEMPLATE_DECL
&& is_specialization_of (supplicant,
TREE_VALUE (friends)))
return 1;
}
break;
}
}
}
else
/* It's a type. */
{
// ...
}
if (declp && DECL_FUNCTION_MEMBER_P (supplicant))
context = DECL_CLASS_CONTEXT (supplicant);
else if (! declp)
/* Local classes have the same access as the enclosing function. */
context = hack_decl_function_context (TYPE_MAIN_DECL (supplicant));
else
context = NULL_TREE;
/* A namespace is not friend to anybody. */
if (context && TREE_CODE (context) == NAMESPACE_DECL)
context = NULL_TREE;
if (context)
return is_friend (type, context);
return 0;
}
基本上,它会获取特定类型的朋友列表,并对其进行迭代,检查其中是否有任何一个与正在测试的函数相等。
通过使用在同一源文件中定义的类似函数将友函数添加到类型中:add_friend()、add_friends() 用于类的所有成员函数,make_friend_class() 用于类等。
我猜它使用这个函数来确定,当检查访问时(它可能会因为访问私有成员而对你大喊大叫的那一刻),该函数是否具有访问权限。
您的问题的答案是(至少对于这个旧版本的 GCC):编译器对每个类都有一个朋友列表,并迭代这个列表以确定一个函数是否是朋友(在编译时,课程。)。如果是,它只是授予访问权限。没有为此生成特殊代码:这是一个常规函数调用。