Why are the following “Resource temporarily unavailable” fingerprints?

Question

Why does the following code print 'read (): a resource is temporarily unavailable in 80% of cases? This is EAGAIN code that matches WHULD BLOCK, which means there is no data waiting to be read, but the choice returns 1, saying that there is data (tested on Linux):

#include <time.h> #include <unistd.h> #include <stdio.h> #include <string.h> #include <sys/types.h> #include <sys/stat.h> #include <fcntl.h> #include <sys/errno.h> int main(int argc, char** argv) { int fd = open("/dev/lp0", O_RDWR | O_NONBLOCK); int ret = 0; int status = 0; char buffer[1024]; char teststr[] = "This is a test\n"; char XMIT_STATUS_OFFLINE[] = {0x10,0x04,0x02}; char XMIT_STATUS_ERROR[] = {0x10,0x04,0x03}; char XMIT_STATUS_ROLL[] = {0x10,0x04,0x04}; char XMIT_STATUS_SLIP[] = {0x10,0x04,0x05}; fd_set rfds; FD_ZERO( &rfds ); FD_SET( fd, &rfds ); struct timeval sleep; sleep.tv_sec = 5; sleep.tv_usec = 0; /* Offline status */ ret = write(fd, XMIT_STATUS_OFFLINE, sizeof(XMIT_STATUS_OFFLINE)); //printf("write() returned %d\n", ret); do { ret = select( fd + 1, &rfds, NULL, NULL, &sleep ); } while (ret < 0 && (errno == EINTR)); ret = read(fd, buffer, 1024); if(ret == -1) { perror("read(): "); } else { status = buffer[0]; if((status & 0x04) != 0) { printf("The cover is open.\n"); } else { printf("OFFLINE is good.\n"); } } close(fd); return 0; } 

Answer 1

Your selected call will return 0 after a 5 second timeout if no data is available. Your code will ignore this and try to read from the device anyway. Check ret == 0 and this will fix your problem.