私有模板类/结构体可见性 [英] Private template classes/structs visibility
问题描述
我不明白为什么在下面的代码中,我允许创建函数 print_private_template
,同时编译器抱怨 print_private_class
:
#include< cstdio>
class A
{
private:
template< unsigned T>
struct B
{
};
struct C
{
};
public:
template< unsigned T>
B< T> getAb()
{
return B< T>();
}
C getAc()
{
return C();
}
};
template< unsigned T>
void print_private_template(const A :: B< T& ab)
{
printf(%d\\\
,T);
}
void print_private_class(const A :: C& ac)
{
printf(something\\\
);
}
int main(int,char **)
{
A a;
print_private_template(a.getAb< 42>());
print_private_class(a.getAc());
return 0;
}
这是一个预期的行为吗?编译器错误/扩展?
为了清楚起见,我的目标是使编译器错误 c> print_private_template 和 print_private_class
。
Comeau 发生错误(当您注释掉 print_private_class $
ComeauTest.c(31):error:class template基于$ b $在实例化print_private_template期间检测到的A :: B(在第7行声明)是不可访问的
void print_private_template(const A :: B& ab)
^ b在第45行的模板参数< 42U>
Windows上的G ++ 4.5不报告 c
A :: B< T>
都具有与任何其他正常成员相同的可见性。因此, print_private_class
和 print_private_template
需要诊断。 < >
11.8嵌套类 [class.access.nest]
1 类是成员,因此具有与任何其他成员相同的访问权限。
的一个封闭类的成员对嵌套类的成员没有特殊访问权限;应遵守通常的访问规则(第11条)
。
I don't understand why in the following code, I am allowed to create the function print_private_template
while the compiler complains about print_private_class
:
#include <cstdio>
class A
{
private:
template <unsigned T>
struct B
{
};
struct C
{
};
public:
template <unsigned T>
B<T> getAb()
{
return B<T>();
}
C getAc()
{
return C();
}
};
template<unsigned T>
void print_private_template(const A::B<T> &ab)
{
printf("%d\n", T);
}
void print_private_class(const A::C &ac)
{
printf("something\n");
}
int main(int, char**)
{
A a;
print_private_template(a.getAb<42>());
print_private_class(a.getAc());
return 0;
}
Is this an expected behaviour? a compiler bug/extension?
Just to be clear, my goal is to make the compiler error on both the usage of print_private_template
and print_private_class
.
Comeau does give an error (when you comment out the print_private_class
function and its call in strict C++03 mode.
ComeauTest.c(31): error: class template "A::B" (declared at line 7) is inaccessible void print_private_template(const A::B &ab) ^ detected during instantiation of "print_private_template" based on template argument <42U> at line 45
G++ 4.5 on Windows does not report any error with -std=c++ -Wall -pedantic
though.
Your class A::C
and class template A::B<T>
both have the same visibility as any other normal members. Hence, both print_private_class
and print_private_template
require a diagnostic.
11.8 Nested classes [class.access.nest]
1 A nested class is a member and as such has the same access rights as any other member. The members of an enclosing class have no special access to members of a nested class; the usual access rules (Clause 11) shall be obeyed.
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