addressing modes in coa.

Question

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)

Answer 1

 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   

Comments

sir but how did we get memory address=20?

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@ritus

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

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