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In an 16 bit instruction the size of address field is 7 bits. The computer uses expanding opcode technique.It has 2, two address instructions and 250 one address instruction. How many Zero address instructions can be formulated ?

  1. 5120
  2. 15304
  3. 768
  4. 1024
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total instruction size= 16 bits

so the total no. of encodings possible=2^16=65536  

no of encodings with 2 address instruction=(2)*(2^7)*(2^7) =2^15=32768    (we have used 2^7 twice as it is 2-address instruction, and 7 is the size of the address field given in the question )

no. of encodings with 1-address instruction= 250*2^7=32000

now, no of 0-adress instruction=65536-(32768+32000) =768

so, the answer is (c) 768 

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