A regression would still absolutely fail and even a flaky pass isn't
harmful as it'd fail most of the time across our N system test runs.
Windows has a low resolution timer and CI systems are prone to odd
timing so this just gives more leeway to avoid flakiness.
Fix the faulthandler implementation of faulthandler.register(signal,
chain=True) if the sigaction() function is not available: don't call
the previous signal handler if it's NULL.
This makes tokenizer.c:valid_utf8 match stringlib/codecs.h:decode_utf8.
It also fixes an off-by-one error introduced in 3.10 for the line number when the tokenizer reports bad UTF8.
This removes the unused `name` variable in the block where `ArgumentTypeError` is handled.
`ArgumentTypeError` errors are handled by showing just the string of the exception; unlike `ValueError`, the name (`__name__`) of the function is not included in the error message.
Fixes#96548
This Monty Python reference is of-its-time. It could seem inappropriate in the context of today's sensibilities around mental health.
Automerge-Triggered-By: GH:iritkatriel
This doesn't happen naturally, but is allowed by the ASDL and compiler.
We don't want to change ASDL for backward compatibility reasons
(#57645, #92987)
Converting a large enough `int` to a decimal string raises `ValueError` as expected. However, the raise comes _after_ the quadratic-time base-conversion algorithm has run to completion. For effective DOS prevention, we need some kind of check before entering the quadratic-time loop. Oops! =)
The quick fix: essentially we catch _most_ values that exceed the threshold up front. Those that slip through will still be on the small side (read: sufficiently fast), and will get caught by the existing check so that the limit remains exact.
The justification for the current check. The C code check is:
```c
max_str_digits / (3 * PyLong_SHIFT) <= (size_a - 11) / 10
```
In GitHub markdown math-speak, writing $M$ for `max_str_digits`, $L$ for `PyLong_SHIFT` and $s$ for `size_a`, that check is:
$$\left\lfloor\frac{M}{3L}\right\rfloor \le \left\lfloor\frac{s - 11}{10}\right\rfloor$$
From this it follows that
$$\frac{M}{3L} < \frac{s-1}{10}$$
hence that
$$\frac{L(s-1)}{M} > \frac{10}{3} > \log_2(10).$$
So
$$2^{L(s-1)} > 10^M.$$
But our input integer $a$ satisfies $|a| \ge 2^{L(s-1)}$, so $|a|$ is larger than $10^M$. This shows that we don't accidentally capture anything _below_ the intended limit in the check.
<!-- gh-issue-number: gh-95778 -->
* Issue: gh-95778
<!-- /gh-issue-number -->
Co-authored-by: Gregory P. Smith [Google LLC] <greg@krypto.org>
Mark Shannon
Brandt Bucher
Gregory P. Smith
Kumar Aditya
Nikita Sobolev
Dennis Sweeney
Guido van Rossum
Steve Dower
Zachary Ware
finefoot
philg314
Pablo Galindo Salgado
Victor Stinner
Itamar Ostricher
Vinay Sajip
Michael Droettboom
Charlie Zhao
Raymond Hettinger
Irit Katriel
Jelle Zijlstra
Jonathon Reinhart
Jason R. Coombs
Ned Deily
Shantanu
Pamela Fox
Erlend E. Aasland
Mark Dickinson