CODEWITHSUNDEEP
rebolred-lang
lyl
lyl

Reputation: 293

How to modify each element of a block by "foreach" in red/rebol

I want to modify each element of a block by foreach. I tried like this but failed:

>> a: [3 4 5 6]
== [3 4 5 6]
>> foreach i a [i + 1]
== 7
>> a
== [3 4 5 6] ;; block a is not changed. What I want is [4 5 6 7]

Is there a better way to achieve it?

Upvotes: 0

Views: 199

Answers (3)

sqlab
sqlab

Reputation: 6436

>> help forall
USAGE:
   FORALL 'word body

DESCRIPTION: 
   Evaluates body for all values in a series. 
   FORALL is a native! value.

ARGUMENTS:
   'word        [word!] "Word referring to series to iterate over."
   body         [block!]

use forall

> a: [3 4 5 6]
== [3 4 5 6]
>> forall a [a/1: a/1 + 1]
== 7
>> probe a
[4 5 6 7]
== [4 5 6 7]

Upvotes: 1

iArnold
iArnold

Reputation: 333

As usual, no reply with your foreach.

a: [2 3 4 5]
b: copy []
foreach i a [append b i + 1]

and if you wish you can set a to b now

a: b

The problem with doing this in one step is that you do not have an index you can use here (despite the suggestive letter i, but that is representing the content of each item inside the block).

So now you can choose your favourite solution.

Upvotes: 1

9214
9214

Reputation: 2193

Changes that you made to values do not persist in a block itself. This ties back to your question about call-by-value parameter passing in Rebol and Red: you modify a copy on the stack (passed down to + along with 1), not the actual value slot that sits inside block a.

To achieve what you want, you need to increment integers in-place, without pushing them on the stack. One way to do so is by using forall.

>> block: [1 2 3]
== [1 2 3]
>> also block forall block [block/1: block/1 + 1]
== [2 3 4]

What forall does is setting a word to a series and then incrementally bumping its index:

>> forall block [probe block]
[1 2 3]
[2 3]
[3]

Since it doesn't extract the actual values, you can access them using path notation, and then modify them in place. block/1 always pick the first value on each iteration.

Upvotes: 1

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