Bug somewhere in :put/map/range/lambda functions?

[Tim Chase]

Tried this:

  :put=range(1,10)

and it worked as expected, appending the numbers 1..10 to my file.
Same with:

  :put=map(range(1,10), 'v:val')

which worked as expected.  However using a no-op lambda function:

  :put=map(range(1,10), {v -> v})

somehow gets the numbers 0..9 in the file instead of 1..10

This reduced it to the simplest case for demonstration.  I stumbled
across it attempting FizzBuzz in a one-liner:

:put=map(range(1, 100), {v -> v % 15 ? v % 5 ? v % 3 ? v : 'Fizz' :
'Buzz' : 'FizzBuzz'})

confused why it kept putting 0/FizzBuzz as the first row.

This is 8.2.1558 on FreeBSD in case it matters:

  :ver
  VIM - Vi IMproved 8.2 (2019 Dec 12, compiled Oct  4 2020 13:32:14)
  Included patches: 1-1558

I'm not sure where I'd start trying to troubleshoot this, but thought
I'd at least mention the oddity here.

-tim




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[Yegappan Lakshmanan]

Hi, Tim,

On Sat, Jan 9, 2021 at 11:27 AM Tim Chase <[hidden email]> wrote:

Tried this:

  :put=range(1,10)

and it worked as expected, appending the numbers 1..10 to my file.
Same with:

  :put=map(range(1,10), 'v:val')

which worked as expected.  However using a no-op lambda function:

  :put=map(range(1,10), {v -> v})

somehow gets the numbers 0..9 in the file instead of 1..10

The lambda function is called with two arguments (the index of the current

item in the list and the value of the item). In your example, you are using

the index of the list item instead of the value. You can try using one of

the following:

     :put=map(range(1, 10), {i, v -> v})

or

     :put=map(range(1, 10), {-> v:val})

 

Regards,

Yegappan

This reduced it to the simplest case for demonstration.  I stumbled
across it attempting FizzBuzz in a one-liner:

:put=map(range(1, 100), {v -> v % 15 ? v % 5 ? v % 3 ? v : 'Fizz' :
'Buzz' : 'FizzBuzz'})

confused why it kept putting 0/FizzBuzz as the first row.

This is 8.2.1558 on FreeBSD in case it matters:

  :ver
  VIM - Vi IMproved 8.2 (2019 Dec 12, compiled Oct  4 2020 13:32:14)
  Included patches: 1-1558

I'm not sure where I'd start trying to troubleshoot this, but thought
I'd at least mention the oddity here.

-tim

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[Tim Chase]

On 2021-01-09 11:39, Yegappan Lakshmanan wrote:
> The lambda function is called with two arguments (the index of the
> current item in the list and the value of the item).
>
>      :put=map(range(1, 10), {i, v -> v})

Ah, that makes sense.  I missed that part in the docs:

  If {expr2} is a |Funcref| it is called with two arguments:
      1. The key or the index of the current item.
      2. the value of the current item.
 
and now I know.  Thanks!

-tim



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