Bit shifting a byte by more than 8 bit

Question

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.

Answer 1

In that arithmetic context byteArray[0] is promoted to either int or unsigned int, so the shift is legal and maybe even sensible (I like to deal only with unsigned types when doing bitwise stuff).

6.5.7 Bitwise shift operators

The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand.

And integer promotions:

6.3.1.1

If an int can represent all values of the original type the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.

Answer 2

The unsigned char's are implicitly cast to int's when shifting. Not sure to what type exactly it is cast, I thing that depends on the platform and the compiler. To get what you intend, it is safer to explicitly cast the bytes, that also makes it more portable and the reader immediately sees what you intend to do:

unsigned long int anotherLongInt;

anotherLongInt = ( ((unsigned long)byteArray[0] << 24) 
               + ((unsigned long)byteArray[1] << 16) 
               + ((unsigned long)byteArray[2] << 8) 
               + ((unsigned long)byteArray[3] ) );