Assume you have a byte-addressable machine that uses 32-bit integers and you are storing the hex value 1234 at address 0:Q. If you wanted to increase the hex value to 123456, which byte assignment would be more efficient, big or little endian? Explain your answer.

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Q: excess notation

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A: The answer is given below.

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Q: ite assembly language code for this problem

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A: SUMMARY: - Hence, we discussed all the points.

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Q: please don't use AI to answer this question. hand-written solution please

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A: Actually, binary numbers are nothing but a 0's and 1's.