USART Programming AVR microcontroler

Question

I want to establish an USART communication between my Atmega32a chip and my computer but there's something I don't understand. As I read the datasheet the UBRR register is a 12 bit register but is split into two 8 bit registers:UBRRH which cointains the four most significant bits and the UBRRL which contains the eight least significant bits of the USART baud rate; My question is:Why I have to shift BAUD_PRESCALLER with 8 at right? I don't understand how this all works :( The example if BAUD_PRESCALLER is 120,in binary it would be:0b01111000 so 0b01111000>>8 = 0b00000000; => Am I wrong somewhere?

void USART_init(void)
{
    UBRRH = (uint8_t)(BAUD_PRESCALLER>>8); <--- this line
    UBRRL = (uint8_t)(BAUD_PRESCALLER);  <-- and this line
    UCSRB = (1<<RXEN)|(1<<TXEN);
    UCSRC = (1<<UCSZ0)|(1<<UCSZ1)|(1<<URSEL);
}

I will atach the datasheet for UBRR register:

Answer 1

Assume you want a BAUD_PRESCALER = 1024.

The binary representation of 1024 is: 0b00000100 00000000.

I've put a space after 8 bits so that you can distinguish between the 8 least significant bits (LSBs) from the 8 most significant bits (MSBs). In order to configure the UBRR registers, you have to put the 8 MSBs in the UBRRH register, and the 8LSBs in the UBRRL register. To do this, you can simply typecast BAUD_PRESCALER to a uint8_t to get the LSBs:

UBRRL = (uint8_t)(BAUD_PRESCALLER);

This will assign 0b00000000 to UBRRL, since the 8 MSBs will be discarded due to the typecast. Remember, the typecast to uint8_t will only give you the 8 LSBs of whatever number you give it.

Now, for the UBRRH register, you first need to shift the 8 MSBs into the LSBs in order for the typecast to work correctly. So 0b00000100 00000000 must become 0b00000100, hence the right shift by 8 bits:

UBRRH = (uint8_t)(BAUD_PRESCALLER>>8); //Shifts BAUD_PRESCALLER MSBs into the LSBs