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upstream: In waitfd(), when poll returns early we are subtracting 

the elapsed time from the timeout each loop, so we only want to measure the
elapsed time the poll() in that loop, not since the start of the function.
Spotted by chris.xj.zhu at gmail.com, ok djm@

OpenBSD-Commit-ID: 199df060978ee9aa89b8041a3dfaf1bf7ae8dd7a
This commit is contained in:
dtucker@openbsd.org 2021-01-15 02:32:41 +00:00 committed by Darren Tucker
parent a164862dfa
commit 5339ab369c
1 changed files with 2 additions and 2 deletions
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4
misc.c
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@ -1,4 +1,4 @@
/* $OpenBSD: misc.c,v 1.158 2021/01/11 02:12:57 dtucker Exp $ */
/* $OpenBSD: misc.c,v 1.159 2021/01/15 02:32:41 dtucker Exp $ */
/*
* Copyright (c) 2000 Markus Friedl. All rights reserved.
* Copyright (c) 2005-2020 Damien Miller. All rights reserved.
@ -297,10 +297,10 @@ waitfd(int fd, int *timeoutp, short events)
struct timeval t_start;
int oerrno, r;
monotime_tv(&t_start);
pfd.fd = fd;
pfd.events = events;
for (; *timeoutp >= 0;) {
monotime_tv(&t_start);
r = poll(&pfd, 1, *timeoutp);
oerrno = errno;
ms_subtract_diff(&t_start, timeoutp);