fix: #3579 conditional parsed as optional chaining (#3584)

Author: josdejong

src/expression/parse.js

@@ -322,7 +322,13 @@ export const createParse = /* #__PURE__ */ factory(name, dependencies, ({
     }
 
     // check for delimiters consisting of 2 characters
-    if (c2.length === 2 && DELIMITERS[c2]) {
+    // Special case: the check for '?.' is to prevent a case like 'a?.3:.7' from being interpreted as optional chaining
+    // TODO: refactor the tokenization into some better way to deal with cases like 'a?.3:.7', see https://github.com/josdejong/mathjs/pull/3584
+    if (
+      c2.length === 2 &&
+      DELIMITERS[c2] &&
+      (c2 !== '?.' || !parse.isDigit(state.expression.charAt(state.index + 2)))
+    ) {
       state.tokenType = TOKENTYPE.DELIMITER
       state.token = c2
       next(state)

test/unit-tests/expression/parse.test.js

@@ -2185,6 +2185,10 @@ describe('parse', function () {
       assert.strictEqual(parseAndEval('0 > 0 ? 1 : 0 < 0 ? -1 : 0'), 0)
     })
 
+    it('should parse a conditional operator and not optional chaining when followed by a number', function () {
+      assert.strictEqual(parseAndEval('true?.3:.7'), 0.3)
+    })
+
     it('should lazily evaluate conditional expression a ? b : c', function () {
       const scope = {}
       math.parse('true ? (a = 2) : (b = 2)').compile().evaluate(scope)