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3. Assuming that an ARM-based processor register contains Ox12345678, show the order in which the bytes of this word would be written in memory starting at address 0x00000000 in a machine that uses Little-Endian byte ordering.

User KTB
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The main answer to your question is as follows: In a machine that uses Little-Endian byte ordering, the bytes of a word are written in reverse order in memory. Let's take the example where the ARM-based processor register contains the value Ox12345678. To show the order in which the bytes would be written in memory, we start at address 0x00000000: 1. The least significant byte (LSB) is written first. In this case, it is 0x78. 2. The next byte is the second least significant byte (second LSB). In this case, it is 0x56. 3. The third byte is the second most significant byte (second MSB). In this case, it is 0x34. 4. Finally, the most significant byte (MSB) is written last. In this case, it is 0x12. Therefore, in memory starting at address 0x00000000, the bytes of the word Ox12345678 would be written in the following order: 0x78, 0x56, 0x34, 0x12. I hope this long answer helps clarify the concept of Little-Endian byte ordering. If you have any more questions, feel free to ask!
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