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A computer supports one address and two address instructions. All the Addresses are the memory addresses. Memory size is 1Mbyte. How many one address instructions are possible if it has 240 two addresses instructions? (Assume binary instructions code has 48 bits)
in CO and Architecture by (37 points) | 254 views

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 in this problem  total size of instruction code = 48 bit

  memory address = 20 bit

  in two address mode =    tag + address1 + address2

                                    =  8  + 20 + 20

       given that  240  is two address instruction

remaining  =  2^8 -  240

                 =  256 - 240 = 16

 hence  one address instruction  =  16* 220   =  224   

by Loyal (6.3k points)
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0
what is the meaning of tag and how is it 8?
+1
because total instruction 48 bit  if u substract 2 address it means (48-20-20) = 8 bit
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@ritus

sub a,b ----> in this instruction a and b are memory addresses and sub is opcode ( tag in this solution )

0
ok thank u sir.
0
sir but how did we get memory address=20?
+1

@ritus

Memory Size = 1 MB = 220 B ===> 20 bits required to identify one Byte in this architecture

0
but there will be 23 bit not 20 because  it's 1mbyte

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