In here
When converting from bytes buffer back to unsigned long int:
unsigned long int anotherLongInt;
anotherLongInt = ( (byteArray[0] << 24)
+ (byteArray[1] << 16)
+ (byteArray[2] << 8)
+ (byteArray[3] ) );
where byteArray is declared as unsigned char byteArray[4];
Question:
I thought byteArray[1]
would be just one unsigned char (8 bit). When left-shifting by 16, wouldn't that shift all the meaningful bits out and fill the entire byte with 0? Apparently it is not 8 bit. Perhaps it's shifting the entire byteArray which is a consecutive 4 byte? But I don't see how that works.
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