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A computer has 32 bit instruction and 12 bit address . If there are 250 two address instructions , the no. of one -address instructions can be .....

Plz formulate a generic solution for this with diagram .

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...........

 

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Opcode size = 32 -(12+12)= 8 bits ,given 12 bit address and 32 bit instruction

Number of  2 address instructions possible =2^8 =256 (used =250)

   Therefore available = 256-250 =6

Now, Number of 1 address instructions possible =6*2^12 = 24576

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