英文:
How can I hash a type using the std::hash of its base class?
问题
以下是翻译好的代码部分：
#include <unordered_set>
#include <string>
template <bool B>
struct C : std::string
{
C(std::string s) : std::string(s) {}
};
namespace std {
template <bool B>
struct hash<C<B>>
{
std::size_t operator()(const C<B> &k) const {
return std::hash<std::string>()(static_cast<std::string>(k));
}
};
} // namespace std
int main() {
std::unordered_set<C<false>> s;
std::string c = C<false>("a");
s.insert(c);
return 0;
}
英文:
I have a class template C<B> which inherits from std::string, and I want to make it hashable like a normal std::string. I have written the following code, and it compiles.
I wonder if it is the correct implementation. Since a cast from derived class to base class may truncate memory, I wonder if it will cost too much?
#include <unordered_set>
#include <string>
template <bool B>
struct C : std::string
{
C(std::string s) : std::string(s) {}
};
namespace std {
template <bool B>
struct hash<C<B>>
{
std::size_t operator()(const C<B> &k) const {
return std::hash<std::string>()(static_cast<std::string>(k));
}
};
} // namespace std
int main() {
std::unordered_set<C<false>> s;
std::string c = C<false>("a");
s.insert(c);
return 0;
}
答案1
得分: 6
是的，这个实现会工作，但不需要static_cast<std::string>(k) - 并且会创建一个临时的std::string，你可能想避免这种情况。
如果你改为static_cast<const std::string&&>(k)，它就不会创建临时的std::string。如果你只是这样做也一样：
std::size_t operator()(const C<B>& k) const {
return std::hash<std::string>()(k);
}
英文:
Yes, the implementation will work, but static_cast<std::string>(k) is not needed - and will create a temporary std::string which you probably would like to avoid.
If you instead make it static_cast<const std::string&>(k), it will not create a temporary std::string. Same thing if you simply do:
std::size_t operator()(const C<B>& k) const {
return std::hash<std::string>()(k);
}
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