I’ve found this piece of code used several times (also a similar one where it’s used open()
instead of write()
).
int c = write(fd, &v, sizeof(v)); if (c == -1 && errno != EINTR) { perror("Write to output file"); exit(EXIT_FAILURE); }
Why it is checked if && errno != EINTR
here ?
Looking for errno
on man I found the following text about EINTR
, but even if I visited man 7 signal
that doesn’t enlighten me.
EINTR Interrupted function call (POSIX.1); see signal(7).
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Answer
Many system calls will report the EINTR
error code if a signal occurred while the system call was in progress. No error actually occurred, it’s just reported that way because the system isn’t able to resume the system call automatically. This coding pattern simply retries the system call when this happens, to ignore the interrupt.
For instance, this might happen if the program makes use of alarm()
to run some code asynchronously when a timer runs out. If the timeout occurs while the program is calling write()
, we just want to retry the system call (aka read/write, etc).