Question regarding Interrupt tutorial by Nick Gammon.

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1

My question is regarding the Tutorial 5 in Nick Gammon's forum on Arduino.

I don't quite understand the ISR function used for ADC reading. Mainly, line 12 in this code snippet.

ISR (ADC_vect)
  {
  byte low, high;
  
  // we have to read ADCL first; doing so locks both ADCL
  // and ADCH until ADCH is read.  reading ADCL second would
  // cause the results of each conversion to be discarded,
  // as ADCL and ADCH would be locked when it completed.
  low = ADCL;
  high = ADCH;

  adcReading = (high << 8) | low;
  adcDone = true;  
  }  // end of ADC_vect

"adcReading = (high << 8) | low;". Correct me if I'm wrong, it is collecting the values of both ADCL and ADCH into one int data type. But why is "high" being shifted left by 8 positions and then getting bit-wise OR'd with "low"? Won't we lose the information of the first and last bit as OR operation would output 1 even if one or the other bit of the corresponding positions (of the two Bytes compared) are 0 or 1.

Can someone please explain the "What is it doing?" and "Why is it doing?" of this process. And if possible then "How?" as well, but not as need-to-know right now.

Hope I am not breaking any rules by posting this here

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3

Thanks! That clears it up.

4

Shaurcasm:
Thanks! That clears it up.

But, you have done the following:

byte low = ADCL;
byte high = ADCH;

When you shift the high variable to the left by 8-bit, the content of the variable is supposed to be lost. To prevent this event from being happened, we must tell the compiler (through casting) to expand the 'processing buffer size' to 16-bit and then do the shift. That is to say that --

int adcReading = (int)high<<8;
adcReading = adcReading | (int)low;   //8-bit variable low is now made 16-bit by appending 8 zeros to its left

But, why does the following instruction practically work without any explicit casting? It is because the Arduino 'processing buffer size' is 16-bit by default.

adcReading = high<<8|low;
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