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3 votes
3 votes
An IOSA define two different instruction formats. In the first format 6 bits are used for the opcode, among these of opcode bits pattern, three are used as special pattern. When any of these three patterns appear in the opcode field, the instruction is encoded using second format, and actual opcode is present in another 6 bit field. How many unique op code can this ISA support?

64

253

254

189

Can anybody please explain this question...

1 Answer

4 votes
4 votes

I solved it this way:

For the first instruction format, the opcode is of 6 bits. So, this gives $2^6$ possible combinations. But, 3 of these patterns represent other instruction formats, therefore $2^6 - 3$.

Now for each of these 3 special patterns, the opcode is specified in some other 6 bits of the instruction. So, for each of these 3 instructions, we have $2^6$ opcodes possible, i.e. $3 * 2^6$

This gives us the total answer:

 $2^6 - 3 + 3 * 2^6$

= 61+192

= 253

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