llvm dev - Feb 2009 - [LLVMdev] Some enhancements to ImmutableSet and FoldingSet

On Wed, Feb 11, 2009 at 8:24 PM, Ted Kremenek <kremenek at apple.com>
wrote:
> Actually, neither of these methods are needed for ImmutableSet.
>  ImmutableSet already has an 'isEmpty()' method and I have never
really seen
> a case where "size()" needs to be explicitly calculated.  If you
need size()
> itself, however, this seems like a perfectly valid addition.
I need to check for size() == 1 (in order to test whether a range set
has a single possible value).

On Feb 11, 2009, at 8:14 PM, Ben Laurie wrote:
> On Wed, Feb 11, 2009 at 8:24 PM, Ted Kremenek <kremenek at apple.com>
> wrote:
>> Actually, neither of these methods are needed for ImmutableSet.
>> ImmutableSet already has an 'isEmpty()' method and I have never
>> really seen
>> a case where "size()" needs to be explicitly calculated.  If
you
>> need size()
>> itself, however, this seems like a perfectly valid addition.
>
> I need to check for size() == 1 (in order to test whether a range set
> has a single possible value).
Ah.  If that is the case, we can implement a 'isSingleton()' method  
with constant time performance (i.e., check if we have a root, and  
check that the root has no children).  Would that work?

On Thu, Feb 12, 2009 at 4:47 AM, Ted Kremenek <kremenek at apple.com>
wrote:
> On Feb 11, 2009, at 8:14 PM, Ben Laurie wrote:
>
>> On Wed, Feb 11, 2009 at 8:24 PM, Ted Kremenek <kremenek at
apple.com> wrote:
>>>
>>> Actually, neither of these methods are needed for ImmutableSet.
>>> ImmutableSet already has an 'isEmpty()' method and I have
never really
>>> seen
>>> a case where "size()" needs to be explicitly calculated. 
If you need
>>> size()
>>> itself, however, this seems like a perfectly valid addition.
>>
>> I need to check for size() == 1 (in order to test whether a range set
>> has a single possible value).
>
> Ah.  If that is the case, we can implement a 'isSingleton()' method
with
> constant time performance (i.e., check if we have a root, and check that
the
> root has no children).  Would that work?
Sure would!