Re: set of pointers
Richard Herring wrote:
No. According to 5.9/2 [expr.rel], among other restrictions:
"If two pointers p and q 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 of
p<q, p>q, p<=q, and p>=q are unspecified."
I understand that to mean that if you have two pointers, there's no
way of knowing if the first one will evaluate as "smaller than" the
second one (other than actually testing with "p < q"). In other words,
you can't assume that if you eg. allocated two blocks of memory to two
pointers, the first pointer will evaluate to be "smaller than" the
second one. It can be either way, there's no guarantee.
However, if you have two pointers p and q, and p evaluates to be
"smaller than" q, it will *always* do so. The "p < q" will not give
random results, but it will always give true in this case.
In other words, if the values of p and q don't change, then the value
of "p < q" will never change either. This means that pointers can indeed
be used in a set/map without any problems. You just can't know in which
order they will end up there, that's all.
The paragraph you quoted specifically mentions that two pointers
pointing to elements in the same array have a well defined order. That's
because arrays are defined as being one contiguous blocks of memory, and
a later element in the array is guaranteed have a larger memory location
than an earlier element. For this reason &array[6] will always be
smaller than &array[10]. The same goes for struct/class member variables.