如何使 [std::operator""s] 在命名空间中可见？

Question

#include <chrono>

namespace X
{
using namespace std;
struct A
{
    std::chrono::seconds d = 0s; // ok
};
}

namespace Y
{
struct B
{
    std::chrono::seconds d = 0s; // error
};
}

错误信息是：

error : no matching literal operator for call to 'operator""s' with argument of type 'unsigned long long' or 'const char *', and no matching literal operator template std::chrono::seconds d = 0s;

我的问题是：

我不想 use namespace std; 在 namespace Y；那么，我应该如何使 std::operator""s 在 namespace Y 中可见？

Answer 1

如果你想拥有所有的计时文字，那么你可以使用

using namespace std::chrono_literals;

如果你只想operator""s那么你可以使用

using std::chrono_literals::operator""s;

请注意至少在 coliru gcc issues a warning for the above line but clang does not. To me there should be no warning. I have asked a follow up question about this at Should a using command issue a warning when using a reserved identifier?

Answer 2

tl;dr：使用

using namespace std::string_literals

These operators are declared in the namespace std::literals::string_literals, where both literals and string_literals are inline namespaces. Access to these operators can be gained with using namespace std::literals, using namespace std::string_literals, and using namespace std::literals::string_literals.

来源：https://en.cppreference.com/w/cpp/string/basic_string/operator%22%22s

如何使 [std::operator""s] 在命名空间中可见？

How to make [std::operator""s] visible in a namespace?

c++

visibility

namespaces

user-defined-literals

c++11