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A CPU has $24$-$bit$ instructions. A program starts at address $300$ (in decimal). Which one of the following is a legal program counter (all values in decimal)?

1. $400$
2. $500$
3. $600$
4. $700$

edited | 4.4k views

Option $(C)$. $24$ bits = $3$ bytes instructions. So, PC will have multiples of $3$ in it.
by Active (3.3k points)
edited by
+2
Why would PC increment in steps of Instruction size ?

1. Shouldn't program counter increment in steps of word size ?

2. In CISC machine instructions are not of uniform length, the question does not mention uniform instruction length. so what if few instructions were of few of only two bytes ?
+8
Aren't you reading the question? It says "24 bit" instructions. And PC always increments in instruction size as it is nothing but pointer to instructions.
0
Why wont we consider it bit wise ,is memory by default byte addressable....Plz explain me
+4
PC content would always be $300+(k-1)*3$ where $kth$ instruction we want to fetch. And this is divisible by 3.
0
It is not given that CPU is byte addressable. So, let us consider bit addressable first. Therefore,

24* (something) +300 = options.

So, if we put any value in something (integer), we are unable to satisfy above equation.

This means that the CPU is byte addressable.

Hence, answer is  $Option$  $C$.
0

@Akash Papnai

But why to consider bit addressable?

Its always word/Byte addressable. Isn't it??

​​​​​@Arjun Sir's comment.

+1
@jeet

Actually you're right. We should always consider a byte-addressable CPU. But this doesn't mean that you'll neglect to consider a bit addressable CPU. There are bit-addressable CPUs and they are faster comparatively but it will increase the cost of the system.

In the future, if such kind of question comes, then you should directly apply byte-addressable since modern computers are byte-addressable.
0
But I have never heard of bit addressable.

+1

+1

Thanks.

C) is the ans

size of instruction  =  24/8 = 3 bytes.

Program Counter can shift 3 bytes at a time to jump to next instruction.

So the given options must be divisible by 3. only 600 is satisfied.

by Active (4k points)
0
Can you explain  why you divided 24 by 8?
+1
Converting bit to byte and as a instruction size is 3 bytes each,program counter will shift 3 bytes each time so address will be multiple of 3.
After each instruction cycle the PC needs to be updated to point to the next instruction in memory.

INSTRUCTION SIZE : 24BITS or 3Bytes

Address of next instruction : 303 then 306 and so on.

So it is an AP series and only 600 comes under this series. So answer is C
by Junior (809 points)

Actually the question is incomplete. It should be mentioned in the question that the

CPU is byte-addressible so that the PC will increment byte by byte. Now because the instruction is 3byte, the PC has to increment 3 times.

So the answer is 600, since it is the only option which is multiple of 3.

by (69 points)

When nothing is given, we always have to assume byte addressability.

So, 1 instruction = 3 B.

From mem location 300, load 99 instructions. We'll be on mem location 597, and PC would point to 600.

Or simply, look for multiples of 3.

Option C

by Active (2.1k points)