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authorVictor Stinner <victor.stinner@gmail.com>2017-10-24 23:53:32 (GMT)
committerGitHub <noreply@github.com>2017-10-24 23:53:32 (GMT)
commit850a18e03e8f8309bc8c39adc6e7d51a4568cd9a (patch)
tree56455b89cb152566734d1fc5de5c70029c3baa70 /Modules/_threadmodule.c
parent3557b05c5a7dfd7d97ddfd3b79aefd53d25e5132 (diff)
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bpo-30768: Recompute timeout on interrupted lock (GH-4103)
Fix the pthread+semaphore implementation of PyThread_acquire_lock_timed() when called with timeout > 0 and intr_flag=0: recompute the timeout if sem_timedwait() is interrupted by a signal (EINTR). See also the PEP 475. The pthread implementation of PyThread_acquire_lock() now fails with a fatal error if the timeout is larger than PY_TIMEOUT_MAX, as done in the Windows implementation. The check prevents any risk of overflow in PyThread_acquire_lock(). Add also PY_DWORD_MAX constant.
Diffstat (limited to 'Modules/_threadmodule.c')
-rw-r--r--Modules/_threadmodule.c6
1 files changed, 4 insertions, 2 deletions
diff --git a/Modules/_threadmodule.c b/Modules/_threadmodule.c
index 72df78f..99611ee 100644
--- a/Modules/_threadmodule.c
+++ b/Modules/_threadmodule.c
@@ -1363,9 +1363,11 @@ PyInit__thread(void)
if (m == NULL)
return NULL;
- timeout_max = PY_TIMEOUT_MAX / 1000000;
- time_max = floor(_PyTime_AsSecondsDouble(_PyTime_MAX));
+ timeout_max = (double)PY_TIMEOUT_MAX * 1e-6;
+ time_max = _PyTime_AsSecondsDouble(_PyTime_MAX);
timeout_max = Py_MIN(timeout_max, time_max);
+ /* Round towards minus infinity */
+ timeout_max = floor(timeout_max);
v = PyFloat_FromDouble(timeout_max);
if (!v)