VSCode: Define exclude pattern that keep out/Debug/gen

Using a set of exlude pattern it is possible to exclude all out/ files
except for files in out/Debug/gen. This removes the need for a symlink.

Change-Id: Ie2b410fa4d4fe12eecbd9c37e5d677b813819b63
Reviewed-on: https://chromium-review.googlesource.com/985977
Commit-Queue: Christian Dullweber <dullweber@chromium.org>
Reviewed-by: Lutz Justen <ljusten@chromium.org>
Cr-Commit-Position: refs/heads/master@{#547649}

docs/vscode.md

@@ -524,18 +524,19 @@ Here are some key bindings that are likely to be useful for you:
 #### The `out` folder
 Automatically generated code is put into a subfolder of out/, which means that
 these files are ignored by VS Code (see files.exclude above) and cannot be
-opened e.g. from quick-open (`Ctrl+P`). On Linux, you can create a symlink as a
-work-around:
-```
-  cd ~/chromium/src
-  mkdir _out
-  ln -s ../out/Debug/gen _out/gen
-```
-We picked _out since it is already in .gitignore, so it won't show up in git
-status.
-
-Note: As of version 1.9, VS Code does not support negated glob commands, but
-once it does, you can use
+opened e.g. from quick-open (`Ctrl+P`).
+As of version 1.21, VS Code does not support negated glob commands, but you can
+define a set of exclude pattern to include only out/Debug/gen:
+
+"files.exclude": {
+  // Ignore build output folders. Except out/Debug/gen/
+  "out/[^D]*/": true,
+  "out/Debug/[^g]*": true,
+  "out/Debug/g[^e]*": true,
+  "out_*/**": true,
+},
+
+Once it does, you can use
 ```
 "!out/Debug/gen/**": true
 ```