问题
What would happen if you call read (or write, or both) in two different thread, on the same file descriptor (lets says we are interested about a local file, and a it's a socket file descriptor), without using explicitly a synchronization mechanism?
Read and Write are syscall, so, on a single core CPU, it's probably unlucky that two read would be executed "at the same time". But with multiple cores...
What the linux kernel will do?
And let's be a bit more general : is the behavior always the same for other kernels (like BSDs) ?
Edit : According to the close documentation, we should be sure that the file descriptor isn't used by a syscall in an other thread. So it seams that explicit synchronization would be required before closing a file descriptor (and so, also around read/write if thread that may call it are still running).
回答1:
Any system level (syscall) file descriptor access is thread safe in all mainstream UNIX-like OSes. Though depending on the age they are not necessarily signal safe.
If you call read, write, accept or similar on a file descriptor from two different tasks then the kernel's internal locking mechanism will resolve contention.
For reads each byte may be only read once though and writes will go in any undefined order.
The stdio library functions fread, fwrite and co. also have by default internal locking on the control structures, though by using flags it is possible to disable that.
回答2:
The comment about close is because it doesn't make a lot of sense to close a file descriptor in any situation in which some other thread might be trying to use it. So while it is 'safe' as far as the kernel is concerned, it can lead to odd, hard to diagnose corner cases.
If a thread closes a file descriptor while a second thread is trying to read from it, the second thread may get an unexpected EBADF error. Worse, if a third thread is simultaneously opening a new file, that might reallocate the same fd, and the second thread might accidentally read from the new file rather than the one it was expecting...
回答3:
Have a care for those who follow in your footsteps
It's perfectly normal to protect the file descriptor with a mutex semaphore. It removes any dependence on kernel behaviour so your message boundaries are now certain. You then don't have to cite the last paragraph at the bottom of a 15,489 line manpage which explains why the mutex isn't necessary (I exaggerated, but you get my meaning)
It also makes it clear to anyone reading your code that the file descriptor is being used by more than one thread.
Fringe Benefit
There is a fringe benefit to using a mutex that way. Suppose you've got different messages coming from the different threads and some of those messages are more important than others. All you need to do is set the thread priorities to reflect their messages' importance. That way the OS will ensure that your messages will be sent in order of importance for minimal effort on your part.
回答4:
The result would depend on how the threads are scheduled to run at that particular instant in time.
One way to potentially avoid undefined behavior with multi-threading is to assume that you are doing memory operations. E.g. updating a linked list or changing a variable, etc.
If you use mutex/semaphores/lock or some other synchronization mechanism, it should work as intended.
来源：https://stackoverflow.com/questions/17431813/c-read-and-thread-safety-linux