Re: Accented Characters and Counting Syllables

2014-12-08 Thread Nordlöw
On Sunday, 7 December 2014 at 15:47:45 UTC, H. S. Teoh via 
Digitalmars-d-learn wrote:

Ok, thanks.

I just noticed that byGrapheme() lacks bidirectional access. 
Further
it also lacks graphemeStrideBack() in complement to 
graphemeStride()?
Similar to stride() and strideBack(). Is this difficult to 
implement?


Not sure, but I wouldn't be surprised if it is. Unicode 
algorithms are

generally non-trivial.


T


What's the best source of information for these algorithms? Is it 
certain that graphemes iteration is backwards iteratable by 
definition?


Re: Accented Characters and Counting Syllables

2014-12-08 Thread Nordlöw

On Monday, 8 December 2014 at 14:57:06 UTC, Nordlöw wrote:
What's the best source of information for these algorithms? Is 
it certain that graphemes iteration is backwards iteratable by 
definition?


I guess

https://en.wikipedia.org/wiki/Combining_character

could be a good start.


Re: Accented Characters and Counting Syllables

2014-12-07 Thread anonymous via Digitalmars-d-learn

On Saturday, 6 December 2014 at 22:37:19 UTC, Nordlöw wrote:

Given the fact that

static assert(é.length == 2);

I was surprised that

static assert(é.byCodeUnit.length == 2);
static assert(é.byCodePoint.length == 2);


string already iterates over code points. So byCodePoint doesn't
have to do anything on it, and it just returns the same string
again.

string's .length is the number of code units. It's not compatible
with the range primitives. That's why hasLength is false for
string (and wstring). Don't use .length on ranges without
checking hasLength.

So, while é.byCodeUnit and é.byCodePoint have equal
`.length`s, they have different range element counts.


Re: Accented Characters and Counting Syllables

2014-12-07 Thread via Digitalmars-d-learn

On Saturday, 6 December 2014 at 22:37:19 UTC, Nordlöw wrote:

static assert(é.byCodePoint.length == 2);


Huh? Why is byCodePoint.length even defined?


Re: Accented Characters and Counting Syllables

2014-12-07 Thread John Colvin via Digitalmars-d-learn

On Sunday, 7 December 2014 at 13:24:28 UTC, Marc Schütz wrote:

On Saturday, 6 December 2014 at 22:37:19 UTC, Nordlöw wrote:

   static assert(é.byCodePoint.length == 2);


Huh? Why is byCodePoint.length even defined?


because string has ElementType dchar (i.e. it already iterates by 
codepoint), which means that byCodePoint is just the identity 
function.


Re: Accented Characters and Counting Syllables

2014-12-07 Thread via Digitalmars-d-learn

On Sunday, 7 December 2014 at 13:24:28 UTC, Marc Schütz wrote:

On Saturday, 6 December 2014 at 22:37:19 UTC, Nordlöw wrote:

   static assert(é.byCodePoint.length == 2);


Huh? Why is byCodePoint.length even defined?


import std.uni;
pragma(msg, typeof(é.byCodePoint));
= string

Something's very broken...

It's this definition in std.uni:

Range byCodePoint(Range)(Range range)
if(isInputRange!Range  is(Unqual!(ElementType!Range) == 
dchar))

{
return range;
}

`Unqual!(ElementType!string)` is indeed `dchar` because of 
auto-decoding.


Filed as bug:
https://issues.dlang.org/show_bug.cgi?id=13829


Re: Accented Characters and Counting Syllables

2014-12-07 Thread Nordlöw
On Saturday, 6 December 2014 at 23:11:49 UTC, H. S. Teoh via 
Digitalmars-d-learn wrote:

This is a Unicode issue. What you want is neither byCodeUnit nor
byCodePoint, but byGrapheme. A grapheme is the Unicode 
equivalent of
what lay people would call a character. A Unicode character 
(or more
precisely, a code point) is not necessarily a complete 
grapheme, as
your example above shows; it's just a numerical value that 
uniquely

identifies an entry in the Unicode character database.


T


Ok, thanks.

I just noticed that byGrapheme() lacks bidirectional access. 
Further it also lacks graphemeStrideBack() in complement to 
graphemeStride()? Similar to stride() and strideBack(). Is this 
difficult to implement?


Re: Accented Characters and Counting Syllables

2014-12-07 Thread H. S. Teoh via Digitalmars-d-learn
On Sun, Dec 07, 2014 at 02:30:13PM +, Nordlöw via Digitalmars-d-learn 
wrote:
 On Saturday, 6 December 2014 at 23:11:49 UTC, H. S. Teoh via
 Digitalmars-d-learn wrote:
 This is a Unicode issue. What you want is neither byCodeUnit nor
 byCodePoint, but byGrapheme. A grapheme is the Unicode equivalent of
 what lay people would call a character. A Unicode character (or
 more precisely, a code point) is not necessarily a complete
 grapheme, as your example above shows; it's just a numerical value
 that uniquely identifies an entry in the Unicode character database.
 
 
 T
 
 Ok, thanks.
 
 I just noticed that byGrapheme() lacks bidirectional access. Further
 it also lacks graphemeStrideBack() in complement to graphemeStride()?
 Similar to stride() and strideBack(). Is this difficult to implement?

Not sure, but I wouldn't be surprised if it is. Unicode algorithms are
generally non-trivial.


T

-- 
Who told you to swim in Crocodile Lake without life insurance??


Re: Accented Characters and Counting Syllables

2014-12-06 Thread H. S. Teoh via Digitalmars-d-learn
On Sat, Dec 06, 2014 at 10:37:17PM +, Nordlöw via Digitalmars-d-learn 
wrote:
 Given the fact that
 
 static assert(é.length == 2);
 
 I was surprised that
 
 static assert(é.byCodeUnit.length == 2);
 static assert(é.byCodePoint.length == 2);
 
 Isn't there a way to iterate over accented characters (in my case
 UTF-8) in D? Or is this an inherent problem in Unicode? I need this in
 a syllable counting algorithm that needs to distinguish accented and
 non-accented variants of vowels. For example café (2 syllables)
 compared to babe (one syllable.

This is a Unicode issue. What you want is neither byCodeUnit nor
byCodePoint, but byGrapheme. A grapheme is the Unicode equivalent of
what lay people would call a character. A Unicode character (or more
precisely, a code point) is not necessarily a complete grapheme, as
your example above shows; it's just a numerical value that uniquely
identifies an entry in the Unicode character database.


T

-- 
There are 10 kinds of people in the world: those who can count in binary, and 
those who can't.