On Sunday, 2 November 2014 at 15:13:37 UTC, bearophile wrote:
I have often arrays that are sorted, and sometimes I'd like to
append items to them. So I'd like to write something like:
SortedRange!(Foo[], q{ a.x < b.x }) data;
data ~= Foo(5);
immutable n = data.upperBound(Foo(2)).length;
This
On Sunday, 2 November 2014 at 15:13:37 UTC, bearophile wrote:
SortedRange!(Foo[], q{ a.x < b.x }) data;
data ~= Foo(5);
immutable n = data.upperBound(Foo(2)).length;
Have anybody implemented SortedRange? I can't find any refs.
On Sunday, 2 November 2014 at 22:14:33 UTC, Xinok wrote:
In general, I don't feel that SortedRange is an ideal solution
anyways. Wrapping ranges in a struct adds too much overhead and
we can't extend the functionality of it. Would it be possible
to replace SortedRange with a @sorted attribute o
On Sunday, 2 November 2014 at 21:35:00 UTC, Xinok wrote:
Sorry, you're right, it's not an "upper bound". I read that
definition a few times over and still got it wrong. O_o
It is always easier to look for the examples :-).
In terms of what to call it, perhaps what you're looking for is
"upper
On Sunday, 2 November 2014 at 20:19:12 UTC, bearophile wrote:
Xinok:
My concern is that SortedRange only accepts a range which is
random-access and limits its functionality to those
primitives. Concatenation is not required for random-access
ranges, so should we expect SortedRange to overload
On Sunday, 2 November 2014 at 20:06:51 UTC, Ola Fosheim Grøstad
wrote:
On Sunday, 2 November 2014 at 19:54:38 UTC, Xinok wrote:
D got it right. C++ returns an iterator which can be a bit
confusing. D returns a slice so it's meaning is much clearer.
No, according to docs D has defined the low
And while you're at it, consider fixing the name to be accurate.
Call it:
lowerBoundedBy(value)
or something similar.
A lower bound is a single element, not a set.
Xinok:
My concern is that SortedRange only accepts a range which is
random-access and limits its functionality to those primitives.
Concatenation is not required for random-access ranges, so
should we expect SortedRange to overload this operator?
I understand, that's why I am asking this her
On Sunday, 2 November 2014 at 19:54:38 UTC, Xinok wrote:
D got it right. C++ returns an iterator which can be a bit
confusing. D returns a slice so it's meaning is much clearer.
No, according to docs D has defined the lower bound to be the
negation of the lower bound…
https://en.wikipedia
On Sunday, 2 November 2014 at 15:13:37 UTC, bearophile wrote:
I have often arrays that are sorted, and sometimes I'd like to
append items to them. So I'd like to write something like:
SortedRange!(Foo[], q{ a.x < b.x }) data;
data ~= Foo(5);
immutable n = data.upperBound(Foo(2)).length;
This
On Sunday, 2 November 2014 at 17:21:04 UTC, Ola Fosheim Grøstad
wrote:
On Sunday, 2 November 2014 at 16:59:30 UTC, bearophile wrote:
Ola Fosheim Grøstad:
Shouldn't sorted range maintain the invariant automatically
in order to remain typesafe?
Yes, of course.
If SortedRange is fixed, please
On Sunday, 2 November 2014 at 16:59:30 UTC, bearophile wrote:
Ola Fosheim Grøstad:
Shouldn't sorted range maintain the invariant automatically in
order to remain typesafe?
Yes, of course.
If SortedRange is fixed, please also switch the names of
upperBound and lowerBound… They are currently
Ola Fosheim Grøstad:
Shouldn't sorted range maintain the invariant automatically in
order to remain typesafe?
Yes, of course.
Bye,
bearophile
On Sunday, 2 November 2014 at 15:13:37 UTC, bearophile wrote:
This means having an array of Foos as sorted range, and
appending an item to it keeping the sorting invariant (so in
non-release mode the append verifies the array is empty or the
last two items satisfy the sorting invariant), and th
I have often arrays that are sorted, and sometimes I'd like to
append items to them. So I'd like to write something like:
SortedRange!(Foo[], q{ a.x < b.x }) data;
data ~= Foo(5);
immutable n = data.upperBound(Foo(2)).length;
This means having an array of Foos as sorted range, and appending
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