Here's a slightly better stack effect documentation:
`when-zero ( ..a n quot: ( ..a -- ..b ) -- ..b )`
But IIUC the actual stack effect is one of these:
`when-zero ( ..a 0 quot: ( ..a -- ..b ) -- ..b )`
`when-zero ( ..a n#0 quot: ( ..a -- ..b ) -- ..a n )`
where # denotes inequality.
Am I right? How does the compiler handle such ambiguity in the stack effect?
01.03.2017, 19:28, "John Benediktsson" <mrj...@gmail.com>:
A stack effect of ( seq quot -- ) doesn't say what the quot is expected to do...You can play around with it and see how it works:IN: scratchpad "abc" [ "empty" ] when-empty ."abc"IN: scratchpad "" [ "empty" ] when-empty ."empty"We should improve the docs here because they are a little hard to follow, since it refers to ``when-empty`` taking the "first quot" of an ``if-empty``, with an empty second quot.,On Wed, Mar 1, 2017 at 8:18 AM, Alexander Ilin <ajs...@yandex.ru> wrote:Hello!
I'm struggling with some of the Factor's words. Mostly with `when-empty`, `if-zero` and similar. It's difficult for me to understand why they don't work sometimes, while their simpler counterparts (while, if) do work exactly as expected.
Here's the latest installment of my struggle:
IN: scratchpad USE: crypto.xor [ [ empty-xor-key ] when-empty ] infer.
( x -- x )
Why is the stack effect as shown here? The documentation for `when-empty` says it takes `( seq quot -- )`. Therefore, I expect `[ [ empty-xor-key ] when-empty ] infer.` to output `( x -- )`.
When I modify the quotation passed to `when-empty` to be a no-op, I get a compilation error:
IN: scratchpad [ [ 1 drop ] when-empty ] infer.
Error
The input quotations to “if-empty” don't match their expected effects
Input Expected Got
[ 1 drop ] ( ..a -- ..b ) ( -- )
[ ] ( ..a seq -- ..b ) ( -- )
That's completely weird to me. Is the stack effect in the documentation incorrect? Should it be `( seq quot -- seq )`?
And why does it refuse to compile the second example with `1 drop`?
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