Show me how an integer of $32$ bits $0$xAB$341209$ will be stored in the following configuration

little endian byte addressable

big endian byte addressable

little endian word addressable $($a word is $32$ bits$)$

big endian word addressable $($a word is $32$ bits$)$

Hint: the example given in class is word addressable $(16$ bits$)$ big endian and little endian.

One of the main reasons why some math$-$intensive implementation prefers little endianness is because of the ease of carry propagation.

$"$As carry propagation must start at the least significant bit $($and thus byte$),$ multi$-$byte addition can then be carried out with a monotonically$-$incrementing address sequence, a simple operation already present in hardware. On a big$-$endian processor, its addressing unit has to be told how big the addition is going to be so that it can hop forward to the least significant byte."

Please use a picture to illustrate the ease of carry propagation with little endianness