Question

A byte addressable memory contains 0x24 0x4A 0x3E 0x87 0x2A 0x5A 0xA5 0x71 from address 0x10010024. A processor following little endian system, is accessing a word from address 0x10010026. What is the value of the word?(show steps)

Answer 1

first understand what is little endian system

In little endian system, last byte of binary representation of the multibyte data-type is stored first.

so if we are storing 5678 then we are going to store the last bits i.e. of 78 will be stored first

now consider the data given : all the data stored and address is in hexadecimal

Address	value
10010024	24
10010025	4A
10010026	3E
10010027	87
10010028	2A
10010029	5A
1001002a	A5
1001002b	71

a WORD can be of 2 byte , 4 byte etc.

as it is not given about word size so we consider a word to be 4 byte as this is the size of integer

so in little endian system we will store data in binary form and then will convert to decimal value

step1: initially the data is empty

step2: as in little endian the last byte will be accessed first i.e. from address 0x10010026

so data at this address is: 0x3E which is equivalent to 0011 1110

00111110

step3: now the second last byte will be accessed from the next address i.e. from 0x10010027

data at this address is : 0x87 which is : 1000 0111

10000111

00111110

step 4: now the third last byte will be accessed from the next address i.e. from 0x10010028

data at this address is : 0x2A which is 0010 1010

00101010

10000111

00111110

step 5 : now the first byte will be accessed from the next address i.e. from 0x10010029

data at this address is : 5A which is 0101 1010

01011010

00101010

10000111

00111110

this is the final value of the integer which is equivalent to  1512736574

also the value in hexadecimal is 0x5A2A873E

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