computer org.

Question

Consider a system with instruction set that uses a fixed 19 bits instruction length and length of address is 8 bits. There are 6 two address instructions and ‘K’ one address instructions. What is the maximum value of ‘K’ if the number of zero address instruction are 65536?

Comments

256??

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I am also getting K as 256.

Number of zero address instruction = ((2^3 - 6)2^8 - k)2^8

Answer 1

opcode (3 bit)

Address(8 bit)

Address(8 bit)

total no. of instructions = 2^3 = 8

#f two address instructions in use =6

Free opcode = 8-6

#f one address instructions = 2*2^8 =512

#f one address instructions in use = k

Free opcode = 512-k

#f zero address instructions = (512-k)*2^8

(512-k)*2^8  =65536

=> k=256

Comments

free opcode=512-k

but opcode possible are 8...how it comes..plz explain