Need explanation of one line of code

[Fliberty]

I am working through a tutorial trying to learn Python. I have come across a line of code I cannot decipher.

I have asked the developer of the code and their explanation ignored my question. My question seems too basic for them to address.

I have simplified the code to the following:

GPIO.output(22,((j == 1<<i) and GPIO.HIGH or GPIO.LOW))

The index i rotates through 0 to 3

The index j rotates through 0x01, 0x02, 0x03 and 0x04

I know the code sends a "High" or "Low" to GPIO pin 22.

I do not understand the code: (j == 1<<i)
it seems to be checking for equality between a hex number "j" and "1<<i"

Questions:
1. How does this work? In other words, what is it saying?

2. Once evaluated, how does it interact with the following "and" and "or" statement?

3.Is there a simpler way to write this line of code?

[ThiefOfTime]

Hi :)
x << y is a binary operator, which symbolises a left shift. To be a more precise, it shifts the value of x by the value of y to the left: if x is 1 it would be 0001 in binary. so if y is 1 and you have 1 << 1 it would result into 0010 (which is 2 in decimal). If y would be 2 it would result to 0100 (which is 4 in decimal). Though 0x02 is hexadecimal and the result of << would be binary it does not matter. 0x02 results to 2 and so does 1 << 1 which in comparison is True. GPIO.HIGH equals True and GPIO.LOW equals False. So the LED would only work if j == 1 << i is True. The part of the GPIO.HIGH or GPIO.LOW is completely unnecessary. I don't know why he would compare the hex number to the binary number, but your script could be made easier by doing this:

GPIO.output(22,(j == 1<<i))

without knowledge where j and i originate from I can not say how to make it more simple.

[Fliberty]

I had found the << operator but could not imagine why someone would use such complicated code. This is akin to using recursion. . .it is elegant and compact but exceedingly difficult for any subsequent programmer to maintain.

BTW, this is code to run a stepper motor. I am studying other stepper driving scheme to find one I can understand and modify to my own uses.

I just hate giving up on trying to understand something because it is hard.

Thank you for the help.

[ThiefOfTime]

could you show me the original code, maybe I can help you with that :)

[Fliberty]

Sure. Here it is. It is from Frenove "Ultimate Starter Kit for Raspberry Pi"

#!/usr/bin/env python3
########################################################################
# Filename    : SteppingMotor.py
# Description : Drive SteppingMotor
# Author      : www.freenove.com
# modification: 2019/12/27
########################################################################
import RPi.GPIO as GPIO
import time 

motorPins = (12, 16, 18, 22)    # define pins connected to four phase ABCD of stepper motor
CCWStep = (0x01,0x02,0x04,0x08) # define power supply order for rotating anticlockwise 
CWStep = (0x08,0x04,0x02,0x01)  # define power supply order for rotating clockwise

def setup():    
    GPIO.setmode(GPIO.BOARD)       # use PHYSICAL GPIO Numbering
    for pin in motorPins:
        GPIO.setup(pin,GPIO.OUT)
		
# as for four phase stepping motor, four steps is a cycle. the function is used to drive the stepping motor clockwise or anticlockwise to take four steps    
def moveOnePeriod(direction,ms):    
    for j in range(0,4,1):      # cycle for power supply order
        for i in range(0,4,1):  # assign to each pin
            if (direction == 1):# power supply order clockwise
                GPIO.output(motorPins[i],((CCWStep[j] == 1<<i) and GPIO.HIGH or GPIO.LOW))
            else :              # power supply order anticlockwise
                GPIO.output(motorPins[i],((CWStep[j] == 1<<i) and GPIO.HIGH or GPIO.LOW))
        if(ms<3):       # the delay can not be less than 3ms, otherwise it will exceed speed limit of the motor
            ms = 3
        time.sleep(ms*0.001)    
		
# continuous rotation function, the parameter steps specifies the rotation cycles, every four steps is a cycle
def moveSteps(direction, ms, steps):
    for i in range(steps):
        moveOnePeriod(direction, ms)
		
# function used to stop motor
def motorStop():
    for i in range(0,4,1):
        GPIO.output(motorPins[i],GPIO.LOW)
            
def loop():
    while True:
        moveSteps(1,3,512)  # rotating 360 deg clockwise, a total of 2048 steps in a circle, 512 cycles
        time.sleep(0.5)
        moveSteps(0,3,512)  # rotating 360 deg anticlockwise
        time.sleep(0.5)

def destroy():
    GPIO.cleanup()             # Release resource

if __name__ == '__main__':     # Program entrance
    print ('Program is starting...')
    setup()
    try:
        loop()
    except KeyboardInterrupt:  # Press ctrl-c to end the program.
        destroy()

Trying to figure out how to attach the .py file so the indentations remain.

[jefsummers]

If you use the python tags it keeps indentation.
Bit shifting can be useful, there was a posted question here for solving 2^2^34 or something similar. Solution takes less than 1 second by taking 1<<(1<<34). Converting that integer to a string to display takes, well, months.

[Fliberty]

I just read the posts about 2^^2^^34. Impressive how very clever people approach a problem in different manners.

Reminds me of spherical math we had to use in a navigation project for a Coast Guard helicopter. Bit shifting sped everything up tremendously.

BTW: I was the hardware engineer on that project. I designed the board and how to integrate it into the bus. I hired a programmer to do the math. He created the marvelous result.