On macOS, `socket.close()` after `shutdown(SHUT_RDWR)` does not reliably interrupt a blocked `recv()` with a timeout

Bug description:

Thread A is blocked in sock.recv() with a timeout set (sock.settimeout(...)).
Thread B closes the socket by calling sock.shutdown(socket.SHUT_RDWR) immediately followed by sock.close()

In this situation, sometimes thread A's recv() call is interrupted promptly, and sometimes it is not interrupted at all: it blocks for the full timeout duration and only then raises TimeoutError, as if shutdown()/close() had no effect on the already-blocked call.

The flakiness is easy to reproduce with the script below, if you have a Mac.

I have not observed this issue when:

• recv() has no timeout set
• The socket's peer closes its end — but the whole point is I want to close the socket quickly, without depending on the peer's behavior.
• A small delay i.e. time.sleep(0.001)is inserted between sock.shutdown(socket.SHUT_RDWR) and sock.close() (but time.sleep(0) is not enough)

Discovered while investigating https://github.com/python-websockets/websockets/issues/1596.

Reproduction

import socket
import threading
import time

RECV_TIMEOUT = 1  # seconds
TRIALS = 50


def run_trial():
    listener = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    listener.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
    listener.bind(("127.0.0.1", 0))
    listener.listen(1)
    a = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    a.connect(listener.getsockname())
    b, _ = listener.accept()
    listener.close()

    a.settimeout(RECV_TIMEOUT)
    result = {}

    def blocked_recv():
        t0 = time.monotonic()
        try:
            data = a.recv(4096)
            result["outcome"] = f"returned {data!r}"
        except OSError as exc:
            result["outcome"] = f"raised {exc!r}"
        result["elapsed"] = time.monotonic() - t0

    thread = threading.Thread(target=blocked_recv)
    thread.start()
    time.sleep(0.15)  # let the thread actually enter the recv() syscall

    a.shutdown(socket.SHUT_RDWR)
    a.close()

    thread.join()
    b.close()

    return result["elapsed"], result["outcome"]


if __name__ == "__main__":
    hangs = 0
    for i in range(TRIALS):
        elapsed, outcome = run_trial()
        hung = elapsed >= RECV_TIMEOUT * 0.5
        hangs += hung
        print(f"trial {i}: {outcome} after {elapsed:.3f}s {'HUNG' if hung else ''}")
    print(f"\n{hangs}/{TRIALS} trials failed to interrupt recv() promptly")

Expected result

recv() is interrupted within a few milliseconds in every trial, since the socket has been explicitly shut down and closed before the timeout can elapse.

Actual result

A representative run:

27/50 trials failed to interrupt recv() promptly

Individual "HUNG" trials show recv() raising TimeoutError('timed out') after the full RECV_TIMEOUT (1.0s), rather than being interrupted immediately after shutdown()/close() run (~0.15s in, as seen in the non-hung trials).

CPython versions tested on:

3.14

Specifically Python 3.14.4 (main, Apr 18 2026, 07:57:29) [Clang 17.0.0]

Operating systems tested on:

macOS

Specifically macOS 26.5.2 (Darwin 25.5.0), arm64

首先在 macOS 上运行提供的复现，并跟踪 CPython 中的 socket.recv()、socket.shutdown() 和 socket.close() 入口点；该 issue 未指定源文件或回归测试。当 shutdown() 后紧接着执行 close() 时，被阻塞的 recv() 能够及时且可靠地中断，并且已通过测试覆盖所报告的行为，即视为完成。