CODEWITHSUNDEEP
c++c++11templatestypestype-traits
user3882729
user3882729

Reputation: 1534

std::decay and removing const qualifiers

I'm trying to understand the workings of std::decay a little better. Per cppreference it's supposed to remove const and volatile classification from the type as part of other transformations it does. However, below function displays "False", "True" vs. "True", "True" as one might expect. Can someone please clarify why the const is needed when matching against the decayed type here? 

int main()
{
   const char *p = "testing";
   cout << "------------------" << endl;
   cout << boolalpha << is_same<char *, decay_t<decltype(p)>>::value << endl;
   cout << boolalpha << is_same<const char *, decay_t<decltype(p)>>::value << endl;
   cout << "------------------" << endl;
}

Upvotes: 3

Views: 1103

Answers (1)

Ted Lyngmo
Ted Lyngmo

Reputation: 117851

std::decay_t would remove the const from the pointer, had it been const, not the type it's pointing at.

That is, a char* const would decay into a char*.

Upvotes: 5

Related Questions