C ++ 11:私有成员安全 [英] C++11: Private member security
问题描述
让我们考虑下一个代码:
Let's consider the next code:
#include <iostream>
#include "mydemangled.hpp"
using namespace std;
struct A
{
private:
struct B {
int get() const { return 5; }
};
public:
B get() const { return B(); }
};
int main()
{
A a;
A::B b = a.get();
cout << demangled(b) << endl;
cout << b.get() << endl;
}
编译器(gcc 4.7.2) c> A :: B 是私有的。好吧。
所以,我改变代码:
And the compiler (gcc 4.7.2) yells saying that A::B
is private. All right.
So, I change the code:
int main()
{
A a;
cout << demangled(a.get()) << endl;
cout << a.get().get() << endl;
}
且不会叫喊:
$ ./a.out
A::B
5
意思是,我不能创建 A :: B
的实例,但我可以使用它。
所以,新的更改(我的问题的关键)。
Meaning, I can't to create instances of A::B
, but I can use it.
So, new change (the key of my question).
int main()
{
A a;
auto b = a.get();
cout << demangled(b) << endl;
cout << b.get() << endl;
}
并输出:
$ ./a.out
A::B
5
这里有什么麻烦, A :: B
private(因此它的构造函数,复制构造函数等等)?
What is the trouble here, being A::B
private (and thus its constructors, copy constructors and so on)?
推荐答案
通常,访问控制名称或符号,不是
基础实体。有,并且总是,许多
访问私人成员的方法;你不能做的是使用这样的成员的
名称。
In general, access controls names or symbols, not the underlying entities. There are, and always have been, numerous ways of accessing private members; what you cannot do is use the name of such a member.
在你的例子中,你不使用名称,所以没有
问题。
In your examples, you don't use the name, so there is no problem.
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