On macOS, `socket.close()` after `shutdown(SHUT_RDWR)` does not reliably interrupt a blocked `recv()` with a timeout
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- Python
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Descrizione
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 betweensock.shutdown(socket.SHUT_RDWR)andsock.close()(buttime.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
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Direzione di ricerca
Inizia eseguendo la riproduzione fornita su macOS e tracciando i punti di ingresso socket.recv(), socket.shutdown() e socket.close() in CPython; l’issue non indica alcun file sorgente né alcun test di regressione. Il lavoro è completato quando il recv() bloccato viene interrotto tempestivamente e in modo affidabile dopo shutdown() seguito da close(), con una copertura del comportamento segnalato.
Scritto dal modello di indicizzazione a partire dal testo della issue.
Valutazione
- Stack tecnologico
- macos, python
- Ambito
- networking, operating-systems
- Tipo di issue
- Bug
- Difficoltà
- 4/5
- Tempo stimato
- 3-5 giorni
- Stato di attività
- Attiva
- Chiarezza
- Abbastanza chiara
- Idoneità per principianti
- 52/100