Prob with ranges

Ruby
1

i looked at an example
rng.each {| i | block } -> rng

Iterates over the elements rng, passing each in turn to the block.

(10…15).each do |n|
print n, ’ ’
end

produces: 10 11 12 13 14 15

when i try the same, it does not work. here is my attempt.
irb(main):025:0> (15…10).each do |n|
irb(main):026:1* print n
irb(main):027:1> end
=> 15…10
irb(main):028:0> (5…10).each do |n|
irb(main):029:1* print n
irb(main):030:1> end
5678910=> 5…10
irb(main):031:0>

2

On Sat, Oct 18, 2008 at 7:29 PM, Junkone [email protected] wrote:

produces: 10 11 12 13 14 15
irb(main):028:0> (5…10).each do |n|
irb(main):029:1* print n
irb(main):030:1> end
5678910=> 5…10
irb(main):031:0>

yes it does, you just forgot the separator in your call to print

(5…10).each do |n|
print n, ’ ’
end

puts n
will give you a newline.

3

Junkone wrote:

irb(main):025:0> (15…10).each do |n|
irb(main):026:1* print n
irb(main):027:1> end
=> 15…10

That doesn’t work because 15…10 is an empty range.

irb(main):028:0> (5…10).each do |n|
irb(main):029:1* print n
irb(main):030:1> end
5678910=> 5…10

That does work. It did print the numbers right there in front of the =>

HTH,
Sebastian

4

so i cannot have a range that is decreasing in numbers. for eg
1…5 can be 1,2,3,4,5
how can i do 5…1 ie 5,4,3,2,1

On Oct 19, 4:14 am, Sebastian H. [email protected]

5

Junkone wrote:

so i cannot have a range that is decreasing in numbers. for eg
1…5 can be 1,2,3,4,5
how can i do 5…1 ie 5,4,3,2,1

Range is by definition increasing. If you want to iterate over numbers
from 5 to 1, you can do it like this: (1…5).to_a.reverse.each or better
(1…5).to_a.reverse_each (this does not allocate the reversed array,
just iterates in the opposite order). Both these methods allocate the
array of all the values, though, so if your range is big, you should
probably use an explicit loop with a decremented value and a stop
condition.

TPR.

6

On Oct 19, 2008, at 11:04 AM, Junkone wrote:

so i cannot have a range that is decreasing in numbers. for eg
1…5 can be 1,2,3,4,5
how can i do 5…1 ie 5,4,3,2,1

5.downto(1) do |i|
puts i
end

7

Junkone wrote:

so i cannot have a range that is decreasing in numbers. for eg
1…5 can be 1,2,3,4,5
how can i do 5…1 ie 5,4,3,2,1

ri Integer#downto

8

Junkone wrote:

irb(main):025:0> (15…10).each do |n|
irb(main):026:1* print n
irb(main):027:1> end
=> 15…10

I used this as a self-tutorial for learning RSpec for the first time.

% spec ./specification -fs

Synopsis

  • (3…7).to_a #=> [3, 4, 5, 6, 7]
  • (7…3).to_a #=> []
  • reversible(3…7).to_a #=> [3, 4, 5, 6, 7]
  • reversible(7…3).to_a #=> [7, 6, 5, 4, 3]

A forward-pointing Range (3…7) with r.exclude_end? == false

  • has invariant reversible(r) == r
  • has invariant reversible(r).object_id == r.object_id

A backward-pointing Range (7…3) with r.exclude_end? == false

  • has invariant reversible(r) == ReverseRange.new(r.begin, r.end)

A forward-pointing Range (3…7) with r.exclude_end? == true

  • is accepted by reversible()

A backward-pointing Range (7…3) with r.exclude_end? == true

  • is rejected by reversible()

ReverseRange instances are normally created with reversible()

  • reversible(7…3) #=> ReverseRange.new(7, 3)

ReverseRange behaves similarly to Range

  • reversible(7…3).inspect #=> ‘#<ReverseRange 7…3>’
  • reversible(7…3).first #=> 7
  • reversible(7…3).last #=> 3
  • reversible(7…3).include?(9) #=> false
  • reversible(7…3).member?(4) #=> true
  • reversible(7…3).include?(4) #=> true
  • (7…3).include?(4) #=> false
  • has the same case/when semantics as Range

ReverseRange can also be passed to reversible()

  • ReverseRange.new(3, 7).to_a #=> []
  • reversible(ReverseRange.new(3, 7)).to_a #=> [3, 4, 5, 6, 7]
  • reversible(ReverseRange.new(3, 7)).is_a?(Range) #=> true
  • reversible(ReverseRange.new(7, 3)).is_a?(ReverseRange) #=> true

ReverseRange#step(n)

  • argument n must be negative
  • reversible(7…3).to_enum(:step, -1).to_a #=> [7, 6, 5, 4, 3]
  • reversible(7…3).to_enum(:step, -2).to_a #=> [7, 5, 3]
  • reversible(7…3).to_enum(:step, -3).to_a #=> [7, 4]
  • reversible(7…3).to_enum(:step, -4).to_a #=> [7, 3]
  • reversible(7…3).to_enum(:step, -5).to_a #=> [7]
  • reversible(7…3).to_enum(:step, -99).to_a #=> [7]

ReverseRange#step(n) with simple one-character String#pred

  • reversible(‘t’…‘p’).to_enum(:step, -1).to_a #=> %w[t s r q p]
  • reversible(‘t’…‘p’).to_enum(:step, -2).to_a #=> %w[t r p]
  • reversible(‘t’…‘p’).to_enum(:step, -3).to_a #=> %w[t q]
  • reversible(‘t’…‘p’).to_enum(:step, -4).to_a #=> %w[t p]
  • reversible(‘t’…‘p’).to_enum(:step, -5).to_a #=> %w[t]
  • reversible(‘t’…‘p’).to_enum(:step, -99).to_a #=> %w[t]

ReverseRange details

  • has invariant r == r
  • has invariant r == ReverseRange.new(r.begin, r.end)
  • has invariant r.eql?(r) == true
  • has invariant r.eql?(ReverseRange.new(r.begin, r.end)) == true
  • has invariant r.exclude_end? == false
  • has invariant r.inspect == r.to_s
  • calls #pred on elements
  • leaves #pred definition to the user

reversible()

  • accepts Range object
  • accepts Range-quacking object
  • fails if argument does not quack like Range

Finished in 0.029019 seconds

46 examples, 0 failures

http://github.com/quix/reverse_range/tree/master

Past-endpoint ranges (7…3) seemed too ambiguous, so I just disallowed
it. In particular if (3…5).to_a == [3, 4], then
reversible(5…3).to_a == [4, 3]. Do not want.

I considered using the name ‘magnetic’ instead of ‘reversible’, as it
has the effect of flipping around to find the right polarity.

James M. Lawrence

9

James M. Lawrence wrote:

I used this as a self-tutorial for learning RSpec for the first time.

% spec ./specification.rb -fs

Synopsis

  • (3…7).to_a #=> [3, 4, 5, 6, 7]
  • (7…3).to_a #=> []
  • reversible(3…7).to_a #=> [3, 4, 5, 6, 7]
  • reversible(7…3).to_a #=> [7, 6, 5, 4, 3]

[…]

I received an email which indicated this needs some explanation. Yes,
this is the RSpec output. I give it an equation, run it through RSpec,
then transform it as above. So in this case the test and the
documentation are the same thing.