Pointer relational operators do not define a total order (§ 5.9 of the C++11 standard):
If two pointers
p
andq
of the same type point to different objects that are not members of the same object or elements of the same array or to different functions, or if only one of them is null, the results ofp<q
,p>q
,p<=q
, andp>=q
are unspecified.
std::less documentation says:
The partial specialization of
std::less
for any pointer type yields a total order, even if the built-inoperator<
does not.
How does it yield this total order from a partial order?
I am unable to answer to this question by looking at /usr/include/c++/4.9/bits/stl_function.h
for struct less
definitions:
template<typename _Tp = void>
struct less;
template<typename _Tp>
struct less : public binary_function<_Tp, _Tp, bool>
{
bool
operator()(const _Tp& __x, const _Tp& __y) const
{ return __x < __y; }
};
template<>
struct less<void>
{
template <typename _Tp, typename _Up>
auto
operator()(_Tp&& __t, _Up&& __u) const
noexcept(noexcept(std::forward<_Tp>(__t) < std::forward<_Up>(__u)))
-> decltype(std::forward<_Tp>(__t) < std::forward<_Up>(__u))
{ return std::forward<_Tp>(__t) < std::forward<_Up>(__u); }
typedef __is_transparent is_transparent;
};