WiringPi SPI without Chip Select/Enable?

[emaq]

I am wondering how the WiringPi SPI works without using the chip select/enable in the following code.

int wiringPiSPIDataRW (int channel, unsigned char *data, int len)
{
  struct spi_ioc_transfer spi ;

  channel &= 1 ;

  memset (&spi, 0, sizeof (spi)) ;

  spi.tx_buf        = (unsigned long)data ;
  spi.rx_buf        = (unsigned long)data ;
  spi.len           = len ;
  spi.delay_usecs   = spiDelay ;
  spi.speed_hz      = spiSpeeds [channel] ;
  spi.bits_per_word = spiBPW ;

  return ioctl (spiFds [channel], SPI_IOC_MESSAGE(1), &spi) ;
}

It seems if I send two consecutive bytes (16-bit data) with WiringPi it will be transmitted as synchronized with clock and the corresponding 16-bit data will be received. But in that case do I have to pull down the CE pin of the device connected to raspberry pi SPI (spidev0.0) by manually connecting it to the ground permanently?

A similar following SPI code does not use chip select/enable.

int
SPI::transfer(uint8_t *send, uint8_t *recv, unsigned len)
{
    if ((send == nullptr) && (recv == nullptr)) {
        return -EINVAL;
    }

    // set write mode of SPI
    int result = ::ioctl(_fd, SPI_IOC_WR_MODE, &_mode);

    if (result == -1) {
        PX4_ERR("can’t set spi mode");
        return PX4_ERROR;
    }

    spi_ioc_transfer spi_transfer{};

    spi_transfer.tx_buf = (uint64_t)send;
    spi_transfer.rx_buf = (uint64_t)recv;
    spi_transfer.len = len;
    spi_transfer.speed_hz = _frequency;
    spi_transfer.bits_per_word = 8;

    result = ::ioctl(_fd, SPI_IOC_MESSAGE(1), &spi_transfer);

    if (result != (int)len) {
        PX4_ERR("write failed. Reported %d bytes written (%s)", result, strerror(errno));
        return PX4_ERROR;
    }

    return PX4_OK;
}

Can someone please explain how this works?

 

[emaq]

In the above question, I mean how the WiringPi SPI works without using

spi.cs_change = true;

 

[Aussie Susan]

Not using the 'chip select' (or similarly named signals) is a bad idea as far as I am concerned.
It will work but if there is a single glitch on the clock signal then every exchange is out of sync after that.
Susan