Embedded C (SPI Bitbang Code) Left Shift Confusion?

Question

So Im writing a bitbang SPI sample code for AVR (So I can understand using SPI better and appreciate the SPI hardware), and im a bit confused at some things.

From what I've read on SPI, you shift out the LSB first from master....but im not sure how to do that? I mean essentially I want to pop the LSB off and set the new 8 bit binary value to before but with the LSB popped off

(Im doing this in a loop, taking a unsigned char and then wanting to send out the LSB each loop).

This is what I have so far:

unsigned char transmit_byte(unsigned char data_byte) 
{
    int i;
    char ret_data;

    enable_SPI();       /* Set up correct Data Direction Ports */
    _delay_ms(100);
    ss_low();           /* Set SS Low */
    _delay_ms(100);

    /* SCLK Currently holding low, MOSI is low, SS is low */
    for (i = 0; i <= 7; i++) 
    {
        if (0x80 & SPI_PORT)
        {
            mosi_high();
        }
        else 
        {
            mosi_low();
        }
        /* MOSI Value is latched in */
        _delay_ms(10); // Probably unnecessary

        //something here?
    }
}

Would I just right shift it out?, I will then read the MISO line and check for 1 or 0 and set a new character variable, but that's after each iteration.

Why do you read the port if this is the master code? The master is responsible for generating MOSI and clock both, at the same time - they must be on the same port so you write to them at the same time. Of course depending on clock phase - SPI allows data to be sent later in the phase, and also allows different clock polarity. — Lundin, Nov 30 '18 at 10:26
Another thing you must realize is that SPI is full duplex. You receive at the same time as you send - it is really just a glorified shift register connected to one rx data register and one tx data register. So if you want to bit bang SPI, you must read MISO shortly (~100ns) after you set MOSI. — Lundin, Nov 30 '18 at 10:27
`x >>= 1;` is how you "pop the LSB off and set the new 8 bit binary value to before but with the LSB popped off". All the remaining bits get shifted to the right by one spot, and the left-most spot becomes 0. — David Grayson, Nov 30 '18 at 19:32
First, you have it backwards. SPI is traditionally MSB first, as opposed to an asynchronous UART which is traditionally LSB first. — Chris Stratton, Nov 30 '18 at 22:50

score 1 · Accepted Answer · answered Nov 30 '18 at 15:17

there are a number of ways to do this

unsigned char mask;
...
for (mask = 0x01, i = 0; i <= 7; i++, mask<<=1) 
{
   ...
   if(data_out&mask)
   {
       set mosi to 1
   }
   else
   {
       set mosi to 0
   }
}

or

unsigned char data_temp;
...
data_temp = data_out;
for(i = 0; i <= 7 ; i++)
{
   ...
   if(data_temp&1)
   {
      set mosi to 1
   }
   else
   {
      set mosi to 0
   }
}

are the two main styles.

walk the mask through the bits (you can do this in either direction lsb first or msb first), or walk the bits through a mask (works either direction.

You have the masking and shifting figured out for receiving but not sending?

Embedded C (SPI Bitbang Code) Left Shift Confusion?

1 Answers1