Question

An array of 2 two byte integers is stored in big endian machine in byte addresses as shown below. What will be its storage pattern in little endian machine?

Address	Data
0 × 104	78
0 × 103	56
0 × 102	34
0 × 101	12

1.

0 × 104    12

0 × 103    56

0 × 102    34

0 × 101    78

2.

0 × 104    12

0 × 103    34

0 × 102    56

0 × 101    78
3.

0 × 104    56

0 × 103    78

0 × 102    12

0 × 101    34
4.

0 × 104    56

0 × 103    12

0 × 102    78

0 × 101    34

Answer 1

Correct Answer - Option 3 :

0 × 104    56

0 × 103    78

0 × 102    12

0 × 101    34

Concept -

• Little and big endian are two ways of storing multi-byte data-types like int, float, etc.
• In little endian machines, last byte of binary representation of the multi-byte data-type is stored first.
• On the other hand, in big endian machines, first byte of binary representation of the multi-byte data-type is stored first.

 

Explanation -

Given in the question is that each integer is of two bytes. For little endian-big endian, we will determine MSB and LSB of both the integers.

The big endian format,

104	103	102	101
78	56	34	12
MSB-1	LSB-1	MSB-2	LSB-2

 

MSB-1 refers to MSB of 1^st integer.

For little endian, we simply have to reverse the exchange MSB with LSB and vice versa.

104	103	102	101
56	78	12	78
LSB-1	MSB-1	LSB-2	MSB-2

 

Hence, Option 3 is correct.

NOTE

if this was one integer with 4 bytes, then in little endian format, entire sequence would have been reversed.