IITM interview

Question

Lets say we have a 32bit instruction using immediate addressing mode, where the opcode is say 22bits, and u have to store a 32bit operand inside the remaining 10bits of the instruction. How would u do it?

Comments

can we convert it into another format and store?like converting it into hexa format..

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we  can use CISC based processor architecture because we can not increase or decrease instruction length in RISC based processors..

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In my opinion 10 bit should be divided into two parts 2 bits(4 possible codes ) and 8 bits(each code shold contain 8 bits =36 bits) but only needed is 32 bit , so it can be easily accomodated

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@utk203,i did nt get you.

and how 8 bits = 36 bits??

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we have remaining 10 bits we divided into 2 bits and 8 bits

For 2 bits (4 codes ) for each code we have 8 bits to store operand value (we have enough to include 32 bit operand in 8+8+8+8=32 bits)

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How can I make 100 Rupees with 10 Rupees?

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Sir, 

there is a problem given at 40:19 , how to economically using instruction set

https://www.youtube.com/watch?v=p9wxyIx-j-c&index=12&list=PLC5FEB130D1CE6075&spfreload=10

same like that  extending opcode field to +2 bits we can store operand value part wise

like we use LDA to store 16 bit data we store in part wise 

[e.g. we have to store LDA 3000H then 

LDA 00H

LDA 30H]

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@utk0203,i am still not able to understand this.
you are saying that "For 2 bits (4 codes ) for each code we have 8 bits to store operand value (we have enough to include 32 bit operand in 8+8+8+8=36 bits)"
how using just 8 bits,we will store 32 bit operand?we need '4'  8-bits to store 32 bits operand.

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@arjun sir,we will need '10' 10 rupee notes to make 100 rs.
so are u saying like that we need '4' 8-bits to store 32 bits?but how can we do thi??

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yes 4 8 bits will be used to store 32 bits

there was error I edited  *(8+8+8+8=32)

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so,,how will u achieve it?will we take 4 instructions to store 32 bits??

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@Akriti Not possible.

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what is not possible sir?

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We cannot have 32 bit immediate operand using 22 bits.

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alright..so with how many min bits we can acquire that??

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32 bits

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sir plz.. explain???

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utk0203

why r u dividing 10 bits in 2 bits and 8 bits?

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|------------opcode(22bits)--------------|------------remaining(10 bits)----|

Now, how this 10 bits of instruction can accomodate 32 bit operand?
do we have to think 1 address instruction for it?

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Ans given :break into 4 8bit portions, and use the remaining 2 bits in each 10bit operand part to sequence them.

but can anyone explain it is right or wrong?

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@Arjun Sir do u mean question is not right?

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No, question is fine but there is no way of doing it.

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how 32 bit operand be inside 10 bit instruction?

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@srestha , See ,

It will stored as part wise first store least significant 8 bits and so on... [ why 8 bits to store operand because rest 2 bits is used to recognize all four different sets of (8bit slot)]

In our microcontroller lab, 8086 microproc. , we used to store 3 byte instructions part wise , that led me to answer this .

Note: - As Sir said,so this may not be correct . In interview you can try like this they will help you to reach the correct solution .