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comp.lang.ruby

substring: to the end of the string

KONTRA Gergely

11/14/2003 3:49:00 PM

Hi!

Is there any easiest way to get the substring starting from the 3rd
character?

my solution:
str[3..-1] which seems quite odd :-/

What if, intuitive, we can write
str[3..]?

Any smarter solution?

Gergo
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14 Answers

Robert Klemme

11/14/2003 4:48:00 PM

0



"KONTRA Gergely" <kgergely@mlabdial.hit.bme.hu> schrieb im Newsbeitrag
news:20031114154849.GA12384@mlabdial.hit.bme.hu...
> Hi!
>
> Is there any easiest way to get the substring starting from the 3rd
> character?
>
> my solution:
> str[3..-1] which seems quite odd :-/
>
> What if, intuitive, we can write
> str[3..]?
>
> Any smarter solution?

Not as far as I know. You can do str[3,str.length-3] but this isn't
really an improvement unless you value the absence of negative integers in
this expression. :-)

robert

Bernard Delmée

11/15/2003 12:34:00 AM

0

> my solution:
> str[3..-1] which seems quite odd :-/

The idiom in python is str[3:]
Conversely, str[:3] gives you the 3 1st chars.

Read the explanation of slicing operations here:
http://www.python.org/doc/current/lib/typ...
(read note 4 & 5)

And beware, 'str' is a builtin, hence a poor choice as a variable name.

Have fun,

Bernard.

Bernard Delmée

11/15/2003 12:45:00 AM

0

Oops - wrong NG, my bad ;-)

Gavin Sinclair

11/15/2003 1:18:00 AM

0

On Saturday, November 15, 2003, 2:48:53 AM, KONTRA wrote:

> Hi!

> Is there any easiest way to get the substring starting from the 3rd
> character?

> my solution:
> str[3..-1] which seems quite odd :-/

Seems quite even to me :)

Seriously, the ability to negative-index strings and arrays like that
is a breath of fresh air compared to traditional compiled languages.
I see no need to improve on it.

> What if, intuitive, we can write
> str[3..]?

I rather doubt that's going to happen. When you execute str[3..7],
the 3..7 isn'y arbitrary syntax, it's a Range object. If Range
objects were allowed to be unbounded, then fine. So the question
really becomes, should Range objects allow unbounded ranges?

> Any smarter solution?

Write a String#substring method if you like.

Cheers,
Gavin

Rodrigo Bermejo

11/15/2003 1:25:00 AM

0

class String

def substring

/(^...)(.*)\z/ =~ self

return $2

end

end

"123456".substring
= >"456"


Now......

I am just wondering how can I build dinamically the Regexp
I mean ...

"123456".substring(5)
= > "6"

"123456".substring(4)
= > "56"

ideas ?

-r.



KONTRA Gergely wrote:

>Hi!
>
>Is there any easiest way to get the substring starting from the 3rd
>character?
>
>my solution:
>str[3..-1] which seems quite odd :-/
>
>What if, intuitive, we can write
>str[3..]?
>
>Any smarter solution?
>
>Gergo
>
>

--
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Advanced Engineering Center
________________________________________
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Information Technologies.
Special Applications
Dial-comm : *879-0644
Phone :(+52) 442-196-0644

Dan Doel

11/15/2003 2:33:00 AM

0

Some ways:

class String
def substring1(n)
reg = Regexp.compile("(^" + "."*n + ")(.*)\\z")
reg.match(self)[2]
end

def substring2(s, e=-1)
self[s..e]
end
end

p "12345".substring1 1
p "12345".substring1 2
p "12345".substring2 1
p "12345".substring2 2
p "12345".substring2 1, 3

Gavin Sinclair

11/15/2003 2:37:00 AM

0

On Saturday, November 15, 2003, 12:24:42 PM, Rodrigo wrote:

> class String
> def substring
> /(^...)(.*)\z/ =~ self
> return $2
> end
> end

> "123456".substring
> = >"456"

> Now......

> I am just wondering how can I build dinamically the Regexp
> I mean ...

> "123456".substring(5)
> = > "6"

> "123456".substring(4)
= >> "56"

> ideas ?

Regexen are overkill here.

class String
def substring(m, n=-1)
self[m..n]
end
end


Gavin

Linus Sellberg

11/15/2003 11:47:00 AM

0

Gavin Sinclair wrote:
> On Saturday, November 15, 2003, 2:48:53 AM, KONTRA wrote:
> Seriously, the ability to negative-index strings and arrays like that
> is a breath of fresh air compared to traditional compiled languages.
> I see no need to improve on it.

I do, but I have no good solution to the problem.

The problem being that positive indexes go from 0 to n-1, while negative
go from -1 to -n, which could introduce confusion.

But as I said, I know of no good solutions, since -0 is not a good way,
and making the first positive index a 1 is just as bad, for several reasons.

Jon A. Lambert

11/15/2003 8:00:00 PM

0


"KONTRA Gergely" <kgergely@mlabdial.hit.bme.hu> wrote in message news:20031114154849.GA12384@mlabdial.hit.bme.hu...
> Hi!
>
> Is there any easiest way to get the substring starting from the 3rd
> character?
>
> my solution:
> str[3..-1] which seems quite odd :-/
>
> What if, intuitive, we can write
> str[3..]?
>
> Any smarter solution?
>

Actually I find this behavior annoying:

irb(main):015:0> "hello"[3..-1]
=> "lo"
irb(main):016:0> "hello"[4..-1]
=> "o"
irb(main):017:0> "hello"[5..-1]
=> ""
irb(main):018:0> "hello"[6..-1]
=> nil

--
J. Lambert

Josef 'Jupp' Schugt

11/15/2003 9:05:00 PM

0

Hi!

* KONTRA Gergely; 2003-11-14, 23:30 UTC:
> Is there any easiest way to get the substring starting from the 3rd
> character?
>
> my solution:
> str[3..-1] which seems quite odd :-/
>
> What if, intuitive, we can write
> str[3..]?
>
> Any smarter solution?

Besids 'get used to -1'? Well, here's some extension to 'String'. I
did try my best to follow POLS. The intention is that left, right,
mid emulate BASIC's left$, right$ and mid$ while head and tail
without any arguments have the results they should have from a
functional programming point of view.

class String

# 'number' leftmost chars
def left(number = 1)
self[0..number-1]
end

# 'number' rightmost chars
def right(number = 1)
self[-number..-1]
end

# 'number' chars starting at position 'from'
def mid(from, number=1)
self[from..from+number]
end

# chars from beginning to 'position'
def head(position = 0)
self[0..position]
end

# chars following 'position'
def tail(position = 0)
self[position+1..-1]
end

end


Josef 'Jupp' Schugt
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