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Your question is a little unclear but I will attempt to answer. Character coding: The character "0" has the code value 48 because that is what the ASCII standard says is the code for character "0". Other encoding standards have made different choices, though nowaways most of those standards have fallen into disuse. Are you asking about the rationale for ...


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As @gnasher729 says, it is really case-dependent. If your question is for the fundamental course purpose, it is better to figure out the specific implementation of processors (stimulators also) and instructions. There are some ways to analyze it generally, like the number of memory access, but not all instructions follow it either.


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Yes, both counter machines and random access machines are Turing complete. The inclusion of indirect addressing does not expand the range of functions that can be computed - it simply makes programming simpler and improves notional efficiency (although the efficiency of an abstract machine is a rather vague concept).


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Just what makes a processor a 32-bit processor? And what would 64-bit word/32-bit halfword mean in the context? There are processors allowing a single instruction to read more than one "row of memory", with documentation more or less readily available. The DEC VAX is notorious for having quad- and octaword accesses (a word being half the width of the ...


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A supplement to the earlier answers: In modern operating systems, programs are often only partially loaded into RAM. The way it works, roughly, is that the first part of the program that is supposed to run is loaded into RAM, and then it starts running. At that point, from the point of view of the running program, all of it is available. So if the code ...


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An 8-kB (8192 bytes) direct-mapped cache has 16-byte lines. The system has 64-bit addresses numbered from 0 on the right to 63 on the left. Which bits are associated with the offset, index, and tag? Since a cache line is 16-byte long, it can be represented with $log_2 16 = 4 \text{b}$, which is the number of bits required for the offset field. The number ...


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Compare $(9.99 \cdot 10^2) \cdot (9.99 \cdot 10^3) = 9.98001 \cdot 10^6$ and $(1.01 \cdot 10^2) \cdot (1.01 \cdot 10^3) = 1.0201 \cdot 10^5$. One has a higher exponent than the other. So in base 10, you'd check if the product of the two mantissas is ≥ 10, and in this case move the decimal point to the left (99.8001 becomes 9.98001) and increase the ...


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1 word=64 bit=8 B 64 B=8 word machine cycle time=2 microseconds => 1 word is transferred from io buffer to memory by cpu in 2 microseconds => data preparation time by cpu 64 B data and io bandwidth is 1MBps so 64/10^6 seconds = 64 microseconds = transfer time by io device = time to dump data from io device to io buffer now in cycle stealing mode ...


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You are both wrong and right. The wrong part is that floating point numbers are stored binary in computers of today. Your example is in decimal. You are right in that what call mantissa overflow actually happens. Before the return to user program however the FPU does a renormalization. There are a few odd cases of course. You might want to read up on ieee ...


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To complement Erik Eidt's answer, the Flag Register conceptionally gets input from the ALU and is heavily used by the CU. ("The Memory Interface Registers" typically do not feature in the programming model.)


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