shiftOut() using shorts/longs

Using Arduino Programming Questions
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does the function take the most significant bit from the original variable (so always 0 for numbers < 255) or does it take it from only the smallest 8 bits of it (so 1 if the number >127)?

What do you think the third parameter of the function does ?

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jaskamakkara:
Is that how it works?

EDIT - the reason I ask is because shiftOut() only shifts out a single byte, so I am unsure how it deals with shorts and ints etc that are still <255 in magnitude.

the function takes an unsigned 8 bit value, a byte in Arduino.

if you put some other type into the function, it will be implicitly converted to a byte.

if you are using other data types, you should select the 8 bits you want to send...

See bit manipulation

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jaskamakkara:
That would seem to make sense to me, but I can't find a reference that confirms it.

you can confirm it by casting yourself:

void setup() 
{
  Serial.begin(9600);
  uint16_t value = 0xDEAD;
  Serial.println((byte)value, HEX);
}

void loop() 
{
}

output:

AD
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Well, that's literally my question. The MSB of a short/int would be the 16th bit,

That is irrelevant. As you've learned, the function takes a byte. The MSBFIRST value tells the function to shift the most significant bit out first, then the second most significant bit, up to the least significant bit OF THE BYTE.

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Where myData is a short or long variable (but still <= 255 in magnitude),

If you know that that value will be <= 255 then is it necessary to use a larger data type in the first place ?

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