mbus_serial_connect() is ready

This commit is contained in:
jakubovsky 2012-07-04 18:38:50 +02:00
parent c848090f56
commit 645e5eb482

View File

@ -32,65 +32,63 @@
int
mbus_serial_connect(mbus_handle *handle)
{
// mbus_serial_handle *serial_data;
//
// if (device == NULL)
// {
// return NULL;
// }
//
// if ((handle = (mbus_serial_handle *)malloc(sizeof(mbus_serial_handle))) == NULL)
// {
// fprintf(stderr, "%s: failed to allocate memory for handle\n", __PRETTY_FUNCTION__);
// return NULL;
// }
//
// handle->device = device; // strdup?
//
// //
// // create the SERIAL connection
// //
//
// // Use blocking read and handle it by serial port VMIN/VTIME setting
// if ((handle->fd = open(handle->device, O_RDWR | O_NOCTTY)) < 0)
// {
// fprintf(stderr, "%s: failed to open tty.", __PRETTY_FUNCTION__);
// return NULL;
// }
//
// memset(&(handle->t), 0, sizeof(handle->t));
// handle->t.c_cflag |= (CS8|CREAD|CLOCAL);
// handle->t.c_cflag |= PARENB;
//
// // No received data still OK
// handle->t.c_cc[VMIN] = 0;
//
// // Wait at most 0.2 sec.Note that it starts after first received byte!!
// // I.e. if CMIN>0 and there are no data we would still wait forever...
// //
// // The specification mentions link layer response timeout this way:
// // The time structure of various link layer communication types is described in EN60870-5-1. The answer time
// // between the end of a master send telegram and the beginning of the response telegram of the slave shall be
// // between 11 bit times and (330 bit times + 50ms).
// //
// // For 2400Bd this means (330 + 11) / 2400 + 0.05 = 188.75 ms (added 11 bit periods to receive first byte).
// // I.e. timeout of 0.2s seems appropriate for 2400Bd.
//
// handle->t.c_cc[VTIME] = 2; // Timeout in 1/10 sec
//
// cfsetispeed(&(handle->t), B2400);
// cfsetospeed(&(handle->t), B2400);
//
//#ifdef MBUS_SERIAL_DEBUG
// printf("%s: t.c_cflag = %x\n", __PRETTY_FUNCTION__, handle->t.c_cflag);
// printf("%s: t.c_oflag = %x\n", __PRETTY_FUNCTION__, handle->t.c_oflag);
// printf("%s: t.c_iflag = %x\n", __PRETTY_FUNCTION__, handle->t.c_iflag);
// printf("%s: t.c_lflag = %x\n", __PRETTY_FUNCTION__, handle->t.c_lflag);
//#endif
//
// tcsetattr(handle->fd, TCSANOW, &(handle->t));
mbus_serial_handle *serial_data;
const char *device;
struct termios *term;
return 0;
if (handle == NULL)
return 0;
serial_data = (mbus_serial_handle *) handle->auxdata;
if (serial_data == NULL || serial_data->device == NULL)
return 0;
device = serial_data->device;
term = &(serial_data->t);
//
// create the SERIAL connection
//
// Use blocking read and handle it by serial port VMIN/VTIME setting
if ((handle->fd = open(device, O_RDWR | O_NOCTTY)) < 0)
{
fprintf(stderr, "%s: failed to open tty.", __PRETTY_FUNCTION__);
return 0;
}
memset(term, 0, sizeof(*term));
term->c_cflag |= (CS8|CREAD|CLOCAL);
term->c_cflag |= PARENB;
// No received data still OK
term->c_cc[VMIN] = 0;
// Wait at most 0.2 sec.Note that it starts after first received byte!!
// I.e. if CMIN>0 and there are no data we would still wait forever...
//
// The specification mentions link layer response timeout this way:
// The time structure of various link layer communication types is described in EN60870-5-1. The answer time
// between the end of a master send telegram and the beginning of the response telegram of the slave shall be
// between 11 bit times and (330 bit times + 50ms).
//
// For 2400Bd this means (330 + 11) / 2400 + 0.05 = 188.75 ms (added 11 bit periods to receive first byte).
// I.e. timeout of 0.2s seems appropriate for 2400Bd.
term->c_cc[VTIME] = 2; // Timeout in 1/10 sec
cfsetispeed(term, B2400);
cfsetospeed(term, B2400);
#ifdef MBUS_SERIAL_DEBUG
printf("%s: t.c_cflag = %x\n", __PRETTY_FUNCTION__, term->c_cflag);
printf("%s: t.c_oflag = %x\n", __PRETTY_FUNCTION__, term->c_oflag);
printf("%s: t.c_iflag = %x\n", __PRETTY_FUNCTION__, term->c_iflag);
printf("%s: t.c_lflag = %x\n", __PRETTY_FUNCTION__, term->c_lflag);
#endif
tcsetattr(handle->fd, TCSANOW, term);
return 1;
}
//------------------------------------------------------------------------------