https://elixir.bootlin.com/linux/v5.19/source/fs/pipe.c#L247
当一个管道满了的时候,它不应该唤醒读者读出数据吗?
/*
* We only wake up writers if the pipe was full when we started
* reading in order to avoid unnecessary wakeups.
*
* But when we do wake up writers, we do so using a sync wakeup
* (W
我已经写了一个代理,它也复制流量。我正在尝试将网络流量复制到副本服务器,该副本服务器应接收所有输入并处理所有请求。但是,只有主服务器上的响应对客户端可见。高级工作流程如下
Thread 1. Take input from client forward it to a pipe in non-blocking way, and to the server
Thread 2. Read from server and send to client
Thread 3. Read from pipe and forward to replica server
Thread 4. Read from
因此,我有一个从父进程到子进程的文件流--而且大多数情况下都很好。但是,当多次快速读取时,使用fget()将返回NULL,错误设置为“资源暂时不可用”。问题是间歇性的--运行执行读取的脚本时,有时会有fget返回空,有时将不会。
有人能帮我阻止这个错误的发生吗?谢谢!
编辑:这是一些代码..。我不知道还有什么代码会有用吗?有很多
// this is the bit that gets a line from the child
if( fgets( line, MAX_LINE_LENGTH, fpin ) == NULL ) {
if( ferror(fpin) ) {
我想尝试一下命名管道,所以我下载了一段代码,并对其进行了修改以进行测试:
fifoname = '/home/foo/pipefifo' # must open same name
def child( ):
pipeout = os.open(fifoname, os.O_NONBLOCK|os.O_WRONLY)
# open fifo pipe file as fd
zzz = 0
while 1:
time.sleep(zzz)
os.write(pipeo
import os
import fcntl
import time
from subprocess import Popen, PIPE
def setNonBlocking(fd):
"""
Set the file description of the given file descriptor to non-blocking.
"""
flags = fcntl.fcntl(fd, fcntl.F_GETFL)
flags = flags | os.O_NONBLOCK
fcntl.fcnt
我在用龙卷风。我希望下面的代码很容易理解,它产生了一个tcproute进程,并将输出发送到websocket的另一端。
class TracerouteHandler(tornado.websocket.WebSocketHandler):
def open(self,ip):
p = subprocess.Popen(['traceroute',ip],stdout=subprocess.PIPE, stderr=subprocess.PIPE)
while p.poll() is None: #running
line=p.stdout.re
我在web上阅读了关于边缘触发epoll函数的文档如下:
1. The file descriptor that represents the read side of a pipe (rfd) is registered on the epoll instance.
2. A pipe writer writes 2 kB of data on the write side of the pipe.
3. A call to epoll_wait(2) is done that will return rfd as a ready file descriptor.
4. The pipe r
在下面的代码之前,我会这样做:
创建一个管道来读取分叉进程的输出
叉()
execv() python脚本
然后,在父进程中,我会:
//set pipes to non-Blocking
File * cout_f = fdopen(cout_pipe[0], "r");
int flags = fcntl(cout_pipe[0], F_GETFL, 0);
fcntl(cout_pipe[0], F_SETFL, flags|O_NONBLOCK);
// read from pipe and send it up through a callbac
有人问我:
下一段代码出了什么问题?如何修复子的进程代码:
#define BUF_SIZE 4096
int my_pipe[2];
pipe(my_pipe);
char buf[BUF_SIZE];
int status = fork();
//Filled buf with message...
if (status == 0) { /* son process */
close(my_pipe[0]);
write(my_pipe[1],buf,BUF_SIZE*sizeof(char));
exit(0);
}
else { /* father
我使用以下代码打开位于'/tmp/xyz‘的命名管道,只读访问:
#!/usr/bin/perl
use strict;
use Fcntl;
...
sysopen(FIFO, "/tmp/xyz", O_RDONLY) or die ("opening named pipe failed: $!\n");
...
问题是,sysopen在调用时挂起。它既不会“死”,也不会在调用执行后执行代码。
ls -la /tmp/xyz的输出:
prw-r--r-- 1 user group 0 Jun 20 11:45 /tmp/xyz
有人知道发生了什么事吗
如何在linux管道上设置写()的超时?
示例代码:
int fd_pipe = open("/run/some/pipe", O_RDWR);
// here i need to set timeout for 3 seconds somehow, if can't write, code will continue...
write(fd_pipe, something, strlen(something));
// continue executing..
谢谢
这似乎是一件很常见的事情,我已经设法教会了自己让它工作所需的一切,除了我现在有一个问题,这是我的疑难解答。
int nonBlockingPOpen(char *const argv[]){
int inpipe;
pid_t pid;
/* open both ends of pipe nonblockingly */
pid = fork();
switch(pid){
case 0: /*child*/
sleep(1); /*child should open after parent h