Question

A sixteen bit signed number, Y = 479 is to be stored in a byte wide memory system. The byte addressable storage for X in "BIG ENDIA N" format, $---- is: The byte addressable storage for Y in "LITTLE ENDIAN" format $---- is:

Answer 1

- Guiven 16 bit signed number Y = 429. h a Y=429 1) convert 429 ent we have in signed Binary is given by following to Binary, 21479 « Remainder 2239 - Division by 2 is done Stepwise 59 dnrdd.NNN 29 I o'. 14 o 1 3 1 .. Binany number = 11101111l ba bits.

In Sider to make it a signed Binary Number of 16 bits . Append os to above Binary number to make it 16 bit signed Binary Neumbeu A 16 Bit signed Birary number Y = 0000000111011111 8 bits make up a Byte Lets divide Y in Bytes ie 8 bits Each. y = 00000001 1101111! - Byte Byte (MSB byte) (LSB Byte) In a little Endian , the LSB Byle is stored first - Format & correspondingly the other Bytes towards MSB Byle In a Big Endian , the MSB Byte is stored first in Memory Ę corresponding the other bytes towards LSB Byle. Y = 0000 0001 11011111 - - MSB Y= 0000 0001 110110 LSB MSB Y = 0000 0001 1010! MEMORY MEMORY Βγτε вете 11011111 00000001 A Byte A+1 Byte 0000 000 H 1111 -> BYTE "BIG ENDIAN" LITTLE ENDIAN"

In the Byte wide memory system, the data is stored in register banks of memory of each 1 byte width.

Here In The memory example , A and A+1 are consecutive Byte addressable memory address considered in the Byte wide Memory System. The bytes of the given number is stored as above diagram in the LITTLE ENDIAN and BIG ENDIAN format.

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