Fix length(null) to return 0 instead of 4

The unary `length` operator special-cased `undefined` and arrays, then
fell through to `String(s).length`. For `null` that became
`String(null)` -> "null" -> 4. Treat `null` as an explicit empty value
with length 0, keeping `undefined` propagating as `undefined`.

Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>

Author: Sander Toonen

packages/expreszo/src/operators/unary/logical.ts

@@ -7,11 +7,16 @@ export function not(a: any): boolean {
   return !a;
 }
 
-export function length(s: any[] | string | number | undefined): number | undefined {
+export function length(s: any[] | string | number | null | undefined): number | undefined {
   if (s === undefined) {
     return undefined;
   }
 
+  // `null` is an explicit empty value — its length is 0, not `String(null).length` (4).
+  if (s === null) {
+    return 0;
+  }
+
   if (Array.isArray(s)) {
     return s.length;
   }

packages/expreszo/test/functions/functions-string.ts

@@ -15,6 +15,11 @@ describe('String Functions TypeScript Test', function () {
       assert.strictEqual(parser.evaluate('length(undefined)'), undefined);
     });
 
+    it('should return 0 if argument is null', function () {
+      const parser = new Parser();
+      assert.strictEqual(parser.evaluate('length(null)'), 0);
+    });
+
     // Note: length(123) returns 3 (number of digits) because there's also a unary operator
     // that handles numbers, so we don't test the error case for non-string argument
   });