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gh-121657: Display correct error message for yield from outside of a … #121680

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Merged
merged 3 commits into from
Jul 13, 2024
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gh-121657: Display correct error message for yield from outside of a … #121680

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gh-121657: Display correct error message for yield from outside of a …
gege-hoho Jul 13, 2024
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📜🤖 Added by blurb_it.
blurb-it[bot] Jul 13, 2024
5ea0d5f
Update Misc/NEWS.d/next/Core and Builtins/2024-07-13-12-27-31.gh-issu…
gege-hoho Jul 13, 2024
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5 changes: 5 additions & 0 deletions Lib/test/test_generators.py
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Original file line number Diff line number Diff line change
Expand Up @@ -2247,6 +2247,11 @@ def printsolution(self, x):
...
SyntaxError: 'yield' outside function

>>> yield from [1,2]
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Maybe add a test with lambda as well since they are "almost like" functions (but not from a compiler PoV).

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I'll do it in a seperate PR

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I added another test as requested in #121722

Traceback (most recent call last):
...
SyntaxError: 'yield from' outside function

>>> def f(): x = yield = y
Traceback (most recent call last):
...
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2 changes: 2 additions & 0 deletions Misc/NEWS.d/next/Core and Builtins/2024-07-13-12-27-31.gh-issue-121657.wgOYLw.rst
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@@ -0,0 +1,2 @@
Improve the :exc:`SyntaxError` message if the user tries to use
:keyword:`yield from <yield>` outside a function.
2 changes: 1 addition & 1 deletion Python/compile.c
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Expand Up @@ -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");
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