Cobol and python it's some other let me think
The answer is true because it solvers by Parameters
Answer:
Adresses Big endian Little Endian
0x100 AB 56
0x101 12 34
0x102 34 12
0x103 56 AB
0x104 00 76
0x105 2B F8
0x106 F8 2B
0x107 76 00
0x108 00 A1
0x109 08 B0
0x10A B0 08
0x10B A1 00
0x10C 00 01
0x10D 00 00
0x10E 00 00
0x10F 01 00
0x110 FE 34
0x111 DC 12
0x112 12 DC
0x113 34 FE
Explanation:
As, it is mentioned that all the integers have 32 bits and all the data given in the question is in Hexadecimal. In Hexadecimal each number consists of 4 bits.
Each memory location contains 1 byte = 8 bits of data. So we store 2 digits of Hexadecimal number each location from each number in given data. If any number contains less than 32 bits in binary, we will add zeros to complete the number in 32 bits.
In big endian order, data is stored from upper most bit to lower most bit in sequential order and in Little Endian order it is vise versa.
It's one single cycle of a loop
