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Re: [PATCH] tpm: fix crash when FD >= 1024
From: |
Michael Tokarev |
Subject: |
Re: [PATCH] tpm: fix crash when FD >= 1024 |
Date: |
Mon, 11 Sep 2023 14:45:01 +0300 |
User-agent: |
Mozilla/5.0 (X11; Linux x86_64; rv:102.0) Gecko/20100101 Thunderbird/102.15.0 |
11.09.2023 14:36, marcandre.lureau@redhat.com:
From: Marc-André Lureau <marcandre.lureau@redhat.com>
Replace select() with poll() to fix a crash when QEMU has a large number
of FDs.
Fixes:
https://bugzilla.redhat.com/show_bug.cgi?id=2020133
Signed-off-by: Marc-André Lureau <marcandre.lureau@redhat.com>
---
backends/tpm/tpm_util.c | 12 +++---------
1 file changed, 3 insertions(+), 9 deletions(-)
diff --git a/backends/tpm/tpm_util.c b/backends/tpm/tpm_util.c
index a6e6d3e72f..5f4c9f5b6f 100644
--- a/backends/tpm/tpm_util.c
+++ b/backends/tpm/tpm_util.c
@@ -112,12 +112,9 @@ static int tpm_util_request(int fd,
void *response,
size_t responselen)
{
- fd_set readfds;
+ GPollFD fds[1] = { {.fd = fd, .events = G_IO_IN } };
int n;
- struct timeval tv = {
- .tv_sec = 1,
- .tv_usec = 0,
- };
+ int timeout = 1000;
You don't need a variable for this in poll().
Besides, this is clear in the context of this patch, which
says tv_sec=1. Without this context, it becomes suspicious
and catches an eye: too long timeout?
n = write(fd, request, requestlen);
if (n < 0) {
@@ -127,11 +124,8 @@ static int tpm_util_request(int fd,
return -EFAULT;
}
- FD_ZERO(&readfds);
- FD_SET(fd, &readfds);
-
/* wait for a second */
- n = select(fd + 1, &readfds, NULL, NULL, &tv);
+ n = RETRY_ON_EINTR(g_poll(fds, 1, timeout));
It's much better IMHO to use "1000" directly here, esp. since the
comment says about a second.
if (n != 1) {
return -errno;
}
Other than that,
Reviewed-by: Michael Tokarev <mjt@tls.msk.ru>
/mjt