std :: is_default_constructible与私有访问和朋友功能 [英] std::is_default_constructible with private access and friend function
问题描述
我很惊讶地发现 std :: is_default_constructible
似乎忽略了朋友的访问.当在类中声明一个默认的构造函数私有,然后将一个函数作为朋友时,我希望 std :: is_default_constructible
返回true.
I was surprised to discover that std::is_default_constructible
appears to ignore friend access. When declaring a default constructor private in a class and then friending a function, I'd expect that std::is_default_constructible
would return true.
示例:我在Wandbox上运行了以下命令: https://wandbox.org/使用Clang 5.0.0和C ++ 17下的GCC 7.2.0.
Example: I ran the following on Wandbox: https://wandbox.org/ using Clang 5.0.0 and GCC 7.2.0 under C++17.
#include <type_traits>
#include <cassert>
class PrivateConstructor
{
private:
PrivateConstructor() = default;
friend void doIt();
};
void doIt()
{
bool isConstructible = std::is_default_constructible<PrivateConstructor>::value;
PrivateConstructor value;
assert(isConstructible); // FAILS!
}
int main(int,char**)
{
doIt();
return 0;
}
此代码可以编译,但是断言失败.是在标准中明确定义的,还是可能的编译器错误?
This code compiles but the assertion fails. Is the defined explicitly in the standard or is this a possible compiler bug?
推荐答案
您已将函数 doIt()
声明为该类的朋友,但该函数无法访问私有类成员.取而代之的是,函数 std :: is_deafault_constructible
访问类成员.
You declared the function doIt()
to be friend of the class, but that function does not access the private class members. Instead, the function std::is_deafault_constructible
accesses the class members.
template< class T >
struct is_default_constructible : std::is_constructible<T> {};
正确的方法是将 std :: is_default_constructible
声明为朋友类:
The proper way is declaring std::is_default_constructible
to be a friend class:
friend class is_default_constructible<PrivateConstructor>;
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