What is inject doing here?

ruby-talk
1

Why does this not do what I expect?

irb(main):001:0> RUBY_VERSION
=> "1.9.2"
irb(main):002:0> (0..3).inject {|s, i| a = i%2 == 0 ? 1 : -1; p a}
-1
1
-1
=> -1

I would think the result should be

1
-1
1
-1
=> -1

Todd

2

The first two elements of the range are the first two elements to be used as arguments to
the inject block:

First, s = 0, i = 1:

a = 1 % 2 == 0 ? 1 : -1; p a # a = -1

Then, s = -1, i = 2:

a = -1 % 2 == 0 ? 1 : -1; p a # a = 1

Then, s = 1, i = 3:

a = 3 % 2 == 0 ? 1 : -1; p a # a = -1

And the range is exhausted.

Michael Edgar
adgar@carboni.ca
http://carboni.ca/

···

On Mar 2, 2011, at 9:15 PM, Todd Benson wrote:

Why does this not do what I expect?

irb(main):001:0> RUBY_VERSION
=> "1.9.2"
irb(main):002:0> (0..3).inject {|s, i| a = i%2 == 0 ? 1 : -1; p a}
-1
1
-1
=> -1

I would think the result should be

1
-1
1
-1
=> -1

Todd

3

Why is this done using inject? s is never used in the block, but since `p a`
returns nil, s would be nil in every iteration after the first one.

(0..3).map { |n| n % 2 == 0 ? 1 : -1 }.each { |n| puts n }

You could of course just use each by itself:

(0..3).each do |n|
  if n % 2 == 0
    puts 1
  else
    puts -1
  end
end

···

On Thu, Mar 3, 2011 at 2:15 AM, Todd Benson <caduceass@gmail.com> wrote:

irb(main):002:0> (0..3).inject {|s, i| a = i%2 == 0 ? 1 : -1; p a}
-1
1
-1
=> -1

4

I'm pretty sure he's just trying to understand inject's behavior on ranges,
not looking for refactoring advice.

Also, on Ruby 1.9, Kernel#p returns its arguments.

Michael Edgar
adgar@carboni.ca
http://carboni.ca/

···

On Mar 3, 2011, at 5:42 AM, Adam Prescott wrote:

On Thu, Mar 3, 2011 at 2:15 AM, Todd Benson <caduceass@gmail.com> wrote:

irb(main):002:0> (0..3).inject {|s, i| a = i%2 == 0 ? 1 : -1; p a}
-1
1
-1
=> -1

Why is this done using inject? s is never used in the block, but since `p a`
returns nil, s would be nil in every iteration after the first one.

(0..3).map { |n| n % 2 == 0 ? 1 : -1 }.each { |n| puts n }

You could of course just use each by itself:

(0..3).each do |n|
if n % 2 == 0
   puts 1
else
   puts -1
end
end

5

I forgot about that! :slight_smile: I caught that one recently courtesy of JEG2.

···

On Thu, Mar 3, 2011 at 5:46 PM, Michael Edgar <adgar@carboni.ca> wrote:

Also, on Ruby 1.9, Kernel#p returns its arguments.