Consider the following for loop in 'C' or 'c++' or 'java'
for (i=namespaceEnd - 1; i >= 0; i--)
Please help me with code to do the same functionality in ruby
···
--
Posted via http://www.ruby-forum.com/.
Martin Durai <martin@angleritech.com> writes:
Consider the following for loop in 'C' or 'c++' or 'java'
for (i=namespaceEnd - 1; i >= 0; i--)
Please help me with code to do the same functionality in ruby
--
Posted via http://www.ruby-forum.com/\.
How about this:
(namespace_end - 1).downto(0) { |x| puts x }
Hope that helps,
Carl.
Martin Durai wrote:
Consider the following for loop in 'C' or 'c++' or 'java'
for (i=namespaceEnd - 1; i >= 0; i--)
Please help me with code to do the same functionality in ruby
ruby does not really have a for loop as such but there are various ways
to get the job done. You saw the upto approach. Here is another:
namespaceEnd.times do
# your code in here
end
now, this works for 0..namespace - 1 automatically. If you need to run
in reverse, can use the downto or a calculation.
namespaceEnd.times do |my_var|
some_array[namespaceEnd - my_var] = some_val
end
but that is clunky. If you are iterating through an array you can use
the each loop:
my_arr = ["aaa", "bbb", "ccc", "ddd"]
my_arr.each_with_index {|str, idx| puts "#{idx}. #{str}"}
=>
0. aaa
1. bbb
2. ccc
3. ddd
If you want only the values, use my_arr.each
If you want only the index position, use my_arr.each_index
HTH
···
--
Posted via http://www.ruby-forum.com/\.
Did you actually read a tutorial or book about the language? If not, it's probably easier to do that vs. trying to cover all these basic questions via newsgroup...
Cheers
robert
···
On 21.11.2007 04:59, Martin Durai wrote:
Consider the following for loop in 'C' or 'c++' or 'java'
for (i=namespaceEnd - 1; i >= 0; i--)
Please help me with code to do the same functionality in ruby
"Martin Durai" <martin@angleritech.com> wrote in message
news:f747d23ae4aecdcd871befcbebba0d09@ruby-forum.com...
Consider the following for loop in 'C' or 'c++' or 'java'
for (i=namespaceEnd - 1; i >= 0; i--)
Please help me with code to do the same functionality in ruby
I'm surprised no one's suggested the obvious:
namespace.reverse_each do |i|
# Do something with i
end
...this assumes you don't actually need an index, which is usually the
case. It also assumes that namespaceEnd is the end of a container called
"namespace," or some such...
If you need the index, instead of the array, you'll have to do something
wild and zany, like:
namespace.reverse.each_index do |i|
# Do something with i
end
...does this help?
Thank you carl, iam very new to this language
sorry carl could you help with this code fully
for( int i = namespaceEnd -1; i >= 0; i--) {
if( prefix.equals( namespacePrefix[ i ] ) ) {
return namespaceUri[ i ];
}
thank you in advance
···
--
Posted via http://www.ruby-forum.com/.
Martin Durai wrote:
Thank you carl, iam very new to this language
sorry carl could you help with this code fully
for( int i = namespaceEnd -1; i >= 0; i--) {
if( prefix.equals( namespacePrefix[ i ] ) ) {
return namespaceUri[ i ];
}thank you in advance
(namespace_end - 1).downto(0) { |x|
return namespaceUri if prefix==namespacePrefix }
Todd
···
--
Posted via http://www.ruby-forum.com/\.
Martin Durai <martin@angleritech.com> writes:
Thank you carl, iam very new to this language
sorry carl could you help with this code fully
for( int i = namespaceEnd -1; i >= 0; i--) {
if( prefix.equals( namespacePrefix[ i ] ) ) {
return namespaceUri[ i ];
}thank you in advance
I suspect you're looking for the element in the namespaceUri
array which is at the position determined by looking up the
position of 'prefix' in the namespacePrefix array (how very
unsettling). Assuming namespaceEnd is actually the count of
elements in the namespacePrefix array, would this work?
your_value = (i = namespace_prefix.index(prefix)) ?
namespace_uri[i] :
nil
Where 'your_value' will now contain the namespace_uri value,
or nil if it was not found in namespace_prefix
If i've misunderstood your question, post back with the values
of prefix, namespacePrefix, namespaceUri and namespaceEnd, and
what you expect to get out of it and I'll see if I can't help.
You might consider using an array of namespace objects instead of parallel arrays
class Namespace
attr_reader :prefix, :uri
def initialize(prefix, uri)
@prefix = prefix
@uri = uri
end
end
namespaces = [ Namespace.new("prefix", "uri"), ...]
then your code becomes something like:
matching_namespace = namespaces.find {|namespace| namespace.prefix == prefix)
return matching_namespace ? matching_namespace.uri : nil
···
On Wed, 21 Nov 2007 13:39:58 +0900, Martin Durai <martin@angleritech.com> wrote:
Thank you carl, iam very new to this language
sorry carl could you help with this code fully
for( int i = namespaceEnd -1; i >= 0; i--) {
if( prefix.equals( namespacePrefix[ i ] ) ) {
return namespaceUri[ i ];
}thank you in advance
Todd Burch <promos@burchwoodusa.com> writes:
Martin Durai wrote:
Thank you carl, iam very new to this language
sorry carl could you help with this code fully
for( int i = namespaceEnd -1; i >= 0; i--) {
if( prefix.equals( namespacePrefix[ i ] ) ) {
return namespaceUri[ i ];
}thank you in advance
(namespace_end - 1).downto(0) { |x|
return namespaceUri if prefix==namespacePrefix }Todd
I'm getting a 'LocalJumpError: unexpected return' error when
I try to return from within a downto block, is this supported
in your version?
Carl.
Paul McMahon <pm@ubit.com> writes:
You might consider using an array of namespace objects instead of parallel
arraysclass Namespace
attr_reader :prefix, :uri
def initialize(prefix, uri)
@prefix = prefix
@uri = uri
end
endnamespaces = [ Namespace.new("prefix", "uri"), ...]
then your code becomes something like:
matching_namespace = namespaces.find {|namespace| namespace.prefix ==
prefix)
return matching_namespace ? matching_namespace.uri : nil
Good point.
I suspect in this case a plain old Hash might do wonders to simplify
the problem...
Carl.
Sorry the code is running well in my system. I didnt got any error. I
have checked the code with my applications.
···
--
Posted via http://www.ruby-forum.com/.