Behavior of word function in local macro

[Noah Blake Smith]

Why does method2 produce the "desired" result, whereas method1 does not? I assume it has to do with differences in word-boundary rules, but I cannot find documentation on it. Thanks in advance!

. local nestedList "a b c" "x y z" "s t u"

. local method1 = word("`nestedList'", 2)

. local method2: word 2 of "`nestedList'"

. macro list
_nestedList:    a b c" "x y z" "s t u
_method1:       b
_method2:       x y z

[Andrew Musau]

This is explained in the respective documentations. From

help macro

Macro functions for parsing

word count string
returns the number of words in string. A token is a word (characters separated by spaces) or set of words enclosed in quotes. Do not enclose string in
double quotes because word count will return 1.

word # of string
returns the #th token of string. Do not enclose string in double quotes.

whereas from

help word()

word(s,n)
Description: the nth word in s; missing ("") if n is missing

Positive numbers count words from the beginning of s, and negative numbers count words from the end of s. (1 is the first word in s, and -1 is the
last word in s.) A word is a set of characters that start and terminate with spaces. This is different from a Unicode word, which is a language
unit based on either a set of word-boundary rules or dictionaries for several languages (Chinese, Japanese, and Thai).

So the -word()- function identifies words using spaces and nothing else.

[daniel klein]

I was mostly wrong in my earlier post; now edited.

Still, you probably want to define the local macro as

local nestedList `""a b c" "x y z" "s t u""' // <- note compound double quotes