template< class T >
struct is_const;
std::is_const
is a UnaryTypeTrait.
If T
is a const-qualified type (that is, const, or const volatile), provides the member constant value equal to true. For any other type, value is false.
If the program adds specializations for std::is_const
or std::is_const_v
, the behavior is undefined.
template< class T >
constexpr bool is_const_v = is_const<T>::value;
T
is a const-qualified type, false otherwise
If T is a reference type then is_const<T>::value is always false. The proper way to check a potentially-reference type for constness is to remove the reference: is_const<typename remove_reference<T>::type>.
[edit] Possible implementation [edit] Example#include <type_traits> static_assert(std::is_same_v<const int*, int const*>, "Remember, constness binds tightly inside pointers."); static_assert(!std::is_const_v<int>); static_assert(std::is_const_v<const int>); static_assert(!std::is_const_v<int*>); static_assert(std::is_const_v<int* const>, "Because the pointer itself can't be changed but the int pointed at can."); static_assert(!std::is_const_v<const int*>, "Because the pointer itself can be changed but not the int pointed at."); static_assert(!std::is_const_v<const int&>); static_assert(std::is_const_v<std::remove_reference_t<const int&>>); struct S { void foo() const {} void bar() const {} }; int main() { // A const member function is const in a different way: static_assert(!std::is_const_v<decltype(&S::foo)>, "Because &S::foo is a pointer."); using S_mem_fun_ptr = void(S::*)() const; S_mem_fun_ptr sfp = &S::foo; sfp = &S::bar; // OK, can be re-pointed static_assert(!std::is_const_v<decltype(sfp)>, "Because sfp is the same pointer type and thus can be re-pointed."); const S_mem_fun_ptr csfp = &S::foo; // csfp = &S::bar; // Error static_assert(std::is_const_v<decltype(csfp)>, "Because csfp cannot be re-pointed."); }[edit] See also checks if a type is volatile-qualified
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