L-1.14: Question on Instruction Format | Computer Organization | UGC NTA NET June 2021

This video contains Question on Instruction Format | Computer Organization |

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[aqsakibria]

Ohk... Who else is follwing his video lectures or the playlist and has got monotonous over "hello doston gate smashers m aapka swagat h"..😂😂. Btw... very nicely explained sir... Lots of respect😊

[biswajit]

Guys, many people in the comments section are getting confused with the final answer. I would like to clarify two points. First of all, there's nothing mentioned about the number of bits required for the "addressing mode" in the question. So, just assume that there's no addressing mode used in this case (because we can't just assume any random number of bits for the same). Secondly, it's given that the immediate operand is a "signed integer". It's not given that "we are storing the number in "signed magnitude form". Here, the number is stored in 2's complement form (any computer will store numbers in 2's complement form only). Even if we take signed representation of the number, the final answer is not going to match with any of the given options. Little tricky but that's how it is :)

[anujsingh-ejvb]

with 8 bits in signed representation we can represent numbers between -127 to +127 as we have two representation for 0 i.e. and in Signed number representation. -128 = = (-1)^1 +

[ItsJyoti]

liked before watching... cz it's known to me that your video can't be waste of time... Thanku Sir

[netcracker]

Thanks sir meine bi aise hi kiya tha. Muje believe nahi hota ki m bi iss type ke questions sahi kar sakti hu. But due to your efforts and Content har subject easy sa ban jata h

[RobinSingh-mszt]

Amazing explanation I understand completely thanks sir.

[ravinjhajharia]

Wouldn't be some bit reserved for addressing mode in the instruction?

[mohangouda]

Bhai very knowledgeable Thanks from deep of our Heart

[RohanSingh-ugtv]

Sir, I think answer should be -64 because 1 bit is used for addressing mode.

[suvadipde]

It will be better if you made more videos on problem-solving of computer organization and architecture

[jyotivarshney]

Video is not available of 1 address, 2 address, 3 address how can understand this concept

[daudahmed]

Isn't it necessary to consider the MSB as the address mode bit of length one? In that case, we will be left with a total of 7 bits for the immediate operand, one will be reserved for sign and rest will be 6 bit which can represent a minimum value of -64?

[jawadudinm]

Sir zero adress aur one adress ka link nahi hai na

[MrHiphopper]

Sirjee signed integer ki range to (-2^n-1)-1 to (+2^n-1)-1 hoti
2's complement ki range (-2^n-1) to (+2^n-1)-1 hoti h
So the answer should be -127, isn't it?

[manjuvimal]

superb ..super sir..we use to do mistake on sign bit..thanks...

[kaivalyastoyscollection]

Thanks, please continue on daily basis

[abhijeetsolanki]

Sir please playlist ka sequence sahi kar dijiye.. Samjhne me bhut problem hoti hai.. Thank you

[snehsrivastava]

Sir I have a doubt:
3. Add #45, when this instruction is executed the following happen/s
a) The processor raises an error and requests for one more operand
b) The value stored in memory location 45 is retrieved and one more operand is requested
c) The value 45 gets added to the value on the stack and is pushed onto the stack
d) None of the mentioned


the answer should be d but as per few sites answer is b... this should mean that 45 should be added to the value in accumulator, right??

[gauravofficial]

Sir I have an doube if OPCODE(6 bit)+ Register1(5 bit) + Register2(5 bit) + Signed Bit(1 bit) = 17 bit, then Immediate Operand = 24 - 17 = 7 = 128 = -64 to 63 ....isnt the ansrer is -64 ??

[abhijeetchauhan]

what is this 32 bit long specifies in the question ?