Question

(1 point) Using a diagram/table, please illustrate how the personal record “John Cool, 45, 1980” will be stored in big- and little-endian byte ordering. The personal record includes a name (string), age (integer), and year (integer). The length of a byte is 8, and the length of a word is 32 for the computer.

Answer 1

Solution for the problem is provided below, please comment if any doubts:

The big-endian format:

• It uses the data stored in the computer in such a manner that the highest significant part of the data will be at lower address.
• Here the data to be stored is “John Cool, 45, 1980”.
• The characters will be stored in 8 bit(7-bit ascii and 1 bit parity).
• The integers are represented using 4 bit hexadecimal numbers.
• Each word consists of 4 bytes and each byte will be stored as given below.
• Note: First column is the address of the memory:

ox….0	J	o	h	n
ox….4		C	o	o
0x….8	l	45	19	80
0x….C

The little-endian format:

• Here the numbers will be stored in such a manner that the lowest significant part will be at lower addressed memory and higher significant part will be at higher addressed memory.
• Here the data to be stored is “John Cool, 45, 1980”.
• The characters will be stored in 8 bit(7-bit ascii and 1 bit parity).
• The integers are represented using 4 bit hexadecimal numbers.
• Each word consists of 4 bytes and each byte will be stored as given below.
• Note: First column is the address of the memory:

ox….0	80	19	45	l
ox….4	o	o	C
0x….8	n	h	o	J
0x….C