what state does shiftOut start / end in?

Forum 2005-2010 (read only) Software Development
#1

Shift out has two pins Clock, Data

I am curious what state shiftOut starts in, and ends in.

what I mean is: When you call shiftOut, does it start by lowering the Clock, Data pins, or does it just start in whatever state you gave it? When shiftOut ends, does it end with the Clock and Data pins low?

I am trying to use shiftOut to control a chip, and I need to understand how shiftOut will work in the middle of all my commands. I have looked at the tutorial but I didn't quite pull an answer to my question out of it.

Thanks! Lucas

#2

It leaves the clock low and the data as whatever the last bit was. Here’s the code (also available in ARDUINO/hardware/cores/arduino):

void shiftOut(uint8_t dataPin, uint8_t clockPin, uint8_t bitOrder, byte val)
{
        int i;

        for (i = 0; i < 8; i++)  {
                if (bitOrder == LSBFIRST)
                        digitalWrite(dataPin, !!(val & (1 << i)));
                else      
                        digitalWrite(dataPin, !!(val & (1 << (7 - i))));
                        
                digitalWrite(clockPin, HIGH);
                digitalWrite(clockPin, LOW);            
        }
}
#3

cool - thanks

#4