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For a 32-bit MIPS, an R-type instruction looks like this: 32-bit MIPS R-type instruction

Say that we expanded the MIPS instruction set to contain four times as many instructions. Would the opcode increase from 6-bits to 8-bits?

Some sources say that the opcode would increase, but many are saying that the opcode wouldn't increase.

I think that the opcode would increase since we now need a larger space to address more instructions.

Community
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jshapy8
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  • There are many possible ways to encode new instructions (including, but not limited to, using new `funct` values, or using non-R-type instructions), and all opcodes aren't used as it is. Just how which instructions ought to be encoded depends a lot on what instructions should be added, so your question does not have a unique answer. – Dolda2000 Feb 14 '17 at 05:34
  • Understood. But in general, increasing the opcode bit field seems acceptable. – jshapy8 Feb 14 '17 at 05:39
  • What fields would you reduce in its stead, though? Increasing the instruction size beyond 32 bits is probably not very acceptable. – Dolda2000 Feb 14 '17 at 05:46
  • I would probably reduce either shamt or funct – jshapy8 Feb 14 '17 at 05:48

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In MIPS, an R type instruction is identified with opcode 0, and the operation is encoded in funct. So, on increasing the number of instructions to four times, funct should increase by 2 bits and not opcode

Ray
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