I tried following code:
#include <iostream>
#include <type_traits>
int main() {
std::cout << std::is_const<std::add_const<int*&>::type>::value;
}
and the output is 0. Is this correct behaviour?
References cannot be const
-qualified, so std::add_const
is a no-op for reference types.
It makes sense when you think about it (and is clearly stated in standard library reference material ), but is potentially surprising at first glance.
std::add_const
does not quite do what you think it does:
If T is a reference , function, or top-level const-qualified type, then type shall name the same type as T , otherwise T const.
(From 20.10.7.1 Const-volatile modifications Table 52 in N4140, emphasis mine)
As you can see, std::add_const
does not add a const
for reference types.
But what would a constant reference be in the first place, they are immutable anyways. (Not to be confused with reference to const
vs. reference to non- const
.)
From the std::add_const
documentation :
http://en.cppreference.com/w/cpp/types/add_cv
Provides the member typedef type which is the same as T, except it has a cv-qualifier added (unless T is a function, a reference , or already has this cv-qualifier)
If we check the cpprefernece documentation for std::add_const we see it says:
Provides the member typedef type which is the same as T, except it has a cv-qualifier added ( unless T is a function, a reference, or already has this cv-qualifier )
this is consistent with the draft C++ standard section 20.10.7.1 Const-volatile modifications which says the following for add_const
:
If T is a reference, function, or top-level const-qualified type, then type shall name the same type as T, otherwise T const.
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