Difficulty: Easy
Correct Answer: thrice
Explanation:
Introduction / Context:Left shifts on binary integers multiply by powers of two. On the 8085, arithmetic left shift is not a single instruction, but RAL (rotate left through carry) effectively shifts left by one bit when used carefully, enabling multiplication by 2^n.
Given Data / Assumptions:
Concept / Approach:Each left shift by one bit multiplies the binary number by 2. Therefore, to multiply by 8, we need three successive left shifts. On 8085, three RAL operations can be used in sequence to achieve this, provided we account for carry if needed.
Step-by-Step Solution:
Target factor: 8 = 2^3.One RAL ≈ one left shift → multiply by 2.Three RAL instructions → multiply by 2^3 = 8.Verification / Alternative check:Test with a small value, e.g., A=0000 0011 (3). After three RALs (ignoring overflow), we get 0001 1000 (24) which is 3*8.
Why Other Options Are Wrong:
Once or twice: gives 2x or 4x, not 8x.Four times: gives 16x, not 8x.Common Pitfalls:
Forgetting that RAL rotates through carry; if high bits spill into carry, you must decide how to handle overflow.Final Answer:
thrice
int range of −32768 to +32767)?
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