gh-121657: Display correct error message for yield from outside of a function (GH-121680)

Co-authored-by: blurb-it[bot] <43283697+blurb-it[bot]@users.noreply.github.com>
Co-authored-by: Bénédikt Tran <[email protected]>

Author: 3 people

Lib/test/test_generators.py

@@ -2247,6 +2247,11 @@ def printsolution(self, x):
   ...
 SyntaxError: 'yield' outside function
 
+>>> yield from [1,2]
+Traceback (most recent call last):
+  ...
+SyntaxError: 'yield from' outside function
+
 >>> def f(): x = yield = y
 Traceback (most recent call last):
   ...

Misc/NEWS.d/next/Core and Builtins/2024-07-13-12-27-31.gh-issue-121657.wgOYLw.rst

@@ -0,0 +1,2 @@
+Improve the :exc:`SyntaxError` message if the user tries to use
+:keyword:`yield from <yield>` outside a function.

Python/compile.c

@@ -6119,7 +6119,7 @@ compiler_visit_expr(struct compiler *c, expr_ty e)
         break;
     case YieldFrom_kind:
         if (!_PyST_IsFunctionLike(SYMTABLE_ENTRY(c))) {
-            return compiler_error(c, loc, "'yield' outside function");
+            return compiler_error(c, loc, "'yield from' outside function");
         }
         if (c->u->u_scope_type == COMPILER_SCOPE_ASYNC_FUNCTION) {
             return compiler_error(c, loc, "'yield from' inside async function");