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Understanding Small part of AVR code

Thread starter #1
Hello fellow Electro-Techs, thank you for your time!! I am currently in the process of combining my love for music with my fascination with ucontrollers and am trying to design a midi to cv converter by myself. I am analysing a code and am currently stuck with the following. It is for a ATtiny2313, the idea is that incoming MIDI data in converted via the Tiny, and then sent to a DAC. BITS 5,6, and 7 of PORT B are the outputs to the DAC. BIT 0 of PORT D is the MIDI input. the rest of PORT D is unused (no idea why they are high).


void make_cv() {
static uint8_t dac_value = 0;
dac_value = (last_active_key()-36)<<1;
PORTD &= ~(0b00111100); // Put Port D bit 2,3,4,5 LOW
PORTD |= (dac_value & 0x0F)<<2;
PORTB &= ~(0b11100000); // Put Port B bit 5,6,7 LOW
PORTB |= (dac_value & 0xF0)<<1;

So last_active_key is a function that returns (as expected) the last key played on a monosynth. Why is it subtracted from 36? what is happening there?

and what does << do?

I have attached a Schematic if it helps! :)
 

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Last edited:

wkrug

Active Member
#2
dac_value = (last_active_key()-36)<<1;
I guess the subtracting of 36 is a "Midi Thing".
Look at MiDi documentation to find out what happens with this - I guess that is the lowest transmitted note.
That's a command to shift binary left.
The nuber behind tells You the number of shifts.
The MSB got be lost at this action.
Example:
Original Byte = 0b00000011
Shift by 1 0b00000011<<1 = 0b00000110
Shift by 2 0b00000011<<2 = 0b00001100

The shift also works with 16, 32, 64, ... Bit variables.

I guess You read something wrong in the schematic.
Pins that are signed with the same label are connected.
PD2..5 from ADC are connected with PD2...5 of the Controller.
 
Last edited:

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