How is this "map" working?

Question

Could you explain what the code below is doing:

resp = s3.list_buckets
puts resp.buckets.map(&:name)

My question is specific to map. I am not able to understand how map is being used here. Also, what does (&:name) mean?

I referred map documentation. However, I'm not able to correlate it with the map in the code above. Per the documentation, Map should be followed by a {}, but it is followed by a () in the code above.

In perl context, map will work on an array/list and will return a new array/list. So, it seems to be doing something similar here as well, but I cannot decode that.

Any pointers to documentation would be helpful.

Have a look at 'with map' section of this http://www.potstuck.com/2011/08/06/ruby-symbols-instead-of-blocks/. It explain all about to_proc method of symbol that you are looking for. — maximus ツ, Dec 11 '13 at 05:08
@mu is too short, Thanks for referring the dups. I used `map` keyword while searching and hence did not find 'em. — slayedbylucifer, Dec 11 '13 at 05:11

Santhosh · Accepted Answer · 2013-12-11T05:04:18.237

2

map is an alias for collect

map(&:name) is shortcut for map {|x| x.name }

map expects a block. & calls to_proc on the object, and passes it as a block and Symbol has to_proc implemented. Refer docs for more info

edited Dec 11 '13 at 05:04

answered Dec 11 '13 at 04:57

Santhosh

28,097
9
82
87

I am able to find more relevant documentation after reading "`map` is an alias for `collect`". Thanks. – slayedbylucifer Dec 11 '13 at 05:35

uncutstone · Answer 2 · 2013-12-11T06:12:31.453

1

& of (&:name) means what follows it should be a Proc object and will be converted to a code block.

Since & expects a Proc object, :name will be converted to a Proc object, this is done by calling :name.to_proc.

After :name is converted to a Proc object, & then converts the resulting Proc object to a code block and gives the code block to map.

In summary, there are totally two type conversions occured, one is converting a symbol to a Proc object, the other is converting a Proc object to a code block.

edited Dec 11 '13 at 06:12

answered Dec 11 '13 at 05:03

uncutstone

408
2
11

Thanks for +1. The explanation is a little ambiguous, so I rewrite a new one. @slayedbylucifer – uncutstone Dec 11 '13 at 06:12

How is this "map" working?

2 Answers2