Ensure backward compatible testing of types (#1266)

- Use "type(0)" instead of "v:t_number"
- Use "==" instead of "==#" since we are comparing numbers
- Make use of the "+=" operator
- Update the "CHANGELOG.md" file

The only real issue here is that we should prefer "type(0)" over the
special variable.  I run into this problem enough on older systems
that it frustrates me.  Let's have it fixed!
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lifecrisis 2021-09-20 15:01:21 -04:00 committed by GitHub
parent 7eee457efa
commit e731b84559
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2 changed files with 6 additions and 5 deletions

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@ -5,6 +5,7 @@
- **.PATCH**: Pull Request Title (PR Author) [PR Number](Link to PR)
-->
#### 6.10
- **.15**: Ensure backward compatible testing of types. (lifecrisis) [#1266](https://github.com/preservim/nerdtree/pull/1266)
- **.14**: Replace trim() with a version-compatible alternative. (PhilRunninger) [#1265](https://github.com/preservim/nerdtree/pull/1265)
- **.13**: Change highlighting of bookmarks in the tree. (PhilRunninger) [#1261](https://github.com/preservim/nerdtree/pull/1261)
- **.12**: Answer the question about accessing files over scp or ftp. (PhilRunninger) [#1259](https://github.com/preservim/nerdtree/pull/1259)

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@ -112,18 +112,18 @@ function! nerdtree#compareNodePaths(p1, p2) abort
" Compare chunks upto common length.
" If chunks have different type, the one which has
" integer type is the lesser.
if type(sortKey1[i]) ==# type(sortKey2[i])
if type(sortKey1[i]) == type(sortKey2[i])
if sortKey1[i] <# sortKey2[i]
return - 1
elseif sortKey1[i] ># sortKey2[i]
return 1
endif
elseif type(sortKey1[i]) ==# v:t_number
elseif type(sortKey1[i]) == type(0)
return -1
elseif type(sortKey2[i]) ==# v:t_number
elseif type(sortKey2[i]) == type(0)
return 1
endif
let i = i + 1
let i += 1
endwhile
" Keys are identical upto common length.