Use process.cwd as root dir if all source files are within it

Author: Daniel-Knights

README.md

@@ -27,7 +27,9 @@ In the absence of `typescript.moduleRoot`, the plugin will mirror the method JSD
 
 1. Parse the referenced module for an `@module` tag.
 2. If a tag is found and it has an explicit id, use that.
-3. If a tag is found, but it doesn't have an explicit id, use the file path relative to the **nearest shared parent directory**, and remove the file extension.
+3. If a tag is found, but it doesn't have an explicit id, use the module's file path relative to the root directory, and remove the file extension.
+
+**NOTE:** The root directory will be `process.cwd()` unless any source files are outside it, in which case, the root folder is considered the nearest shared parent directory of all source files.
 
 ## What this plugin does
 

index.js

@@ -35,17 +35,25 @@ const fileNodes = {};
 let implicitModuleRoot;
 
 /**
- * @return {string} The nearest shared parent directory of all source files.
+ * Without explicit module ids, JSDoc will use `process.cwd()` if all source files are within cwd.
+ * If any source files are outside cwd, JSDoc will use the nearest shared parent directory.
+ * @return {string} The implicit root path with which to resolve all module ids against.
  */
 function getImplicitModuleRoot() {
   if (implicitModuleRoot) {
     return implicitModuleRoot;
   }
 
-  if (!env.sourceFiles || env.sourceFiles.length === 0) {
+  if (
+    !env.sourceFiles ||
+    env.sourceFiles.length === 0 ||
+    // If all files are in cwd
+    env.sourceFiles.every((f) => f.startsWith(process.cwd()))
+  ) {
     return process.cwd();
   }
 
+  // Find the nearest shared parent directory
   implicitModuleRoot = path.dirname(env.sourceFiles[0]);
 
   env.sourceFiles.slice(1).forEach((filePath) => {