Computers and Technology, 10.04.2020 02:03 bre644

Assume a computer that has 32-bit integers. Show how each of the following values would be stored sequentially in memory, starting at address 0x100, assuming that each address holds 1 byte. Be sure to extend each value to the appropriate number of bits. You will need to add more rows (addresses) to store all given values.

a) 0xAB123456

b) 0x2BF876

c) 0x8B0A1

d) 0x1

e) 0xFEDC1234

byte order
address big endian little endian
0x100
0x101
0x102
0x103
0x104
0x105
0x106
0x107

Answers: 3

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Answer from: kam3864

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.

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Answer from: Quest

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Assume a computer that has 32-bit integers. Show how each of the following values would be stored se...

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