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Show how the following value would be stored by byte-addressable machines with 32-bit words by filling in the appropriate memory locations, using big endian and then little endian format. Assume each value starts at address Ox10. VALUE: 0xF12 Address-Ox10 Big Endian Little Endian Ox11 0x12 0x13 0x14 Ox15

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Answer #1

VALUE is: 0xF12, if we write this Value in terms of 32 bit word we will get
VALUE is: 0x00 00 0F 12

Little Endian:
The LSB (least significant Byte is stored at Lowest Address)

Address Data
0x10 12
0x11 0F
0x12 00
0x13 00
0x14 00
0x15 00
0x16 00
0x17 00


Big Endian:
The MSB (most significant Byte is stored at Lowest Address)

Address Data
0x10 00
0x11 00
0x12 00
0x13 00
0x14 00
0x15 00
0x16 0F
0x17 12


Thanks, let me know if there is any concern/doubts.

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