以下代码使用clang和MSVC成功编译但无法在GCC 6.1.0中编译.
#include <memory>
template<typename R, typename T, typename... Args>
T* test(R(T::*)(Args...) const)
{
return nullptr;
}
int main()
{
using T = std::shared_ptr<int>;
T* p = test(&T::get);
}
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以下错误消息
prog.cc: In function 'int main()':
prog.cc:13:16: error: invalid conversion from 'std::__shared_ptr<int, (__gnu_cxx::_Lock_policy)2u>*' to 'T* {aka std::shared_ptr<int>*}' [-fpermissive]
T* p = test(&T::get);
~~~~^~~~~~~~~
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问题是libstdc ++是std::shared_ptr通过get从基类继承成员函数实现的std::__shared_ptr.
在C++标准20.8.2.2类模板shared_ptr中,它指定了std :: shared_ptr类的类定义以及该类的所有成员函数.
我的问题是,实现是否必须至少提供标准类中标准中定义的所有公共类成员?是否允许通过继承libstdc ++中实现的基类来提供成员函数?
struct root
{
int i = 0;
};
struct base: root{};
struct derive: base{};
constexpr derive d0;
int main()
{
constexpr auto& r = static_cast<root const&>(d0);
constexpr auto& b = static_cast<base const&>(r);
constexpr auto& d = static_cast<derive const&>(r);
static_assert(d0.i == 0, ""); // ok
static_assert(r.i == 0, ""); // ok
static_assert(b.i == 0, ""); // error in gcc
static_assert(d.i == 0, ""); // ok
}
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Clang接受上面的代码,但是gcc 7.2.0编译错误如下:
prog.cc:17:5: error: non-constant condition for static assertion
static_assert(b.i == 0, "");
^~~~~~~~~~~~~
prog.cc:17:5: …Run Code Online (Sandbox Code Playgroud)