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b886d83c5b
Based on 1 normalized pattern(s): this program is free software you can redistribute it and or modify it under the terms of the gnu general public license as published by the free software foundation version 2 of the license extracted by the scancode license scanner the SPDX license identifier GPL-2.0-only has been chosen to replace the boilerplate/reference in 315 file(s). Signed-off-by: Thomas Gleixner <tglx@linutronix.de> Reviewed-by: Allison Randal <allison@lohutok.net> Reviewed-by: Armijn Hemel <armijn@tjaldur.nl> Cc: linux-spdx@vger.kernel.org Link: https://lkml.kernel.org/r/20190531190115.503150771@linutronix.de Signed-off-by: Greg Kroah-Hartman <gregkh@linuxfoundation.org>
61 lines
1.3 KiB
C
61 lines
1.3 KiB
C
// SPDX-License-Identifier: GPL-2.0-only
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/* Timeout API for single-threaded programs that use blocking
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* syscalls (read/write/send/recv/connect/accept).
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*
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* Copyright (C) 2017 Red Hat, Inc.
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*
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* Author: Stefan Hajnoczi <stefanha@redhat.com>
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*/
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/* Use the following pattern:
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*
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* timeout_begin(TIMEOUT);
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* do {
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* ret = accept(...);
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* timeout_check("accept");
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* } while (ret < 0 && ret == EINTR);
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* timeout_end();
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*/
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#include <stdlib.h>
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#include <stdbool.h>
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#include <unistd.h>
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#include <stdio.h>
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#include "timeout.h"
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static volatile bool timeout;
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/* SIGALRM handler function. Do not use sleep(2), alarm(2), or
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* setitimer(2) while using this API - they may interfere with each
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* other.
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*/
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void sigalrm(int signo)
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{
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timeout = true;
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}
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/* Start a timeout. Call timeout_check() to verify that the timeout hasn't
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* expired. timeout_end() must be called to stop the timeout. Timeouts cannot
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* be nested.
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*/
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void timeout_begin(unsigned int seconds)
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{
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alarm(seconds);
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}
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/* Exit with an error message if the timeout has expired */
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void timeout_check(const char *operation)
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{
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if (timeout) {
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fprintf(stderr, "%s timed out\n", operation);
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exit(EXIT_FAILURE);
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}
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}
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/* Stop a timeout */
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void timeout_end(void)
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{
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alarm(0);
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timeout = false;
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}
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