If the set of MIPS instructions were changed to accommodate 128 records and 4 times more instructions type-I, which would be the largest hexadecimal immediate value that could be supported, keeping the 32-bit instructions?
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1What are records? Do you perhaps mean registers? – Erik Eidt Oct 29 '16 at 17:02
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Assuming you mean registers not records:
Most of what you're asking about is a matter of encodings which are done with bits.
4x more means 2 more bits are needed, because log2(4) = 2, or in other words, 4 = 22.
The current Type-I instructions:
I instructions are converted into machine code words in the following format:
opcode rs rt IMM 6 bits 5 bits 5 bits 16 bits
The current register fields are 5 bits which can accommodate designation of 25 = 32 different registers. To take that to 128 registers, you'd need log2(128) = 7 (also that 27 = 128) bits, meaning 2 more bits than now.
So in total:
4x more opcodes: 2 more bits are needed.
4x more registers: 2 more bits are needed per register field for a total of 4 more bits
That all adds up to 6 more bits required, which would leave 10 bits left for the immediate field instead of the current 16.
10 bits can be used to represent an unsigned range from 0 to 210-1 or from 0 to 1023. Or 10 bits can be used to represent a signed range from -29 to +29-1, or -512 to 511. There are other encodings possible (one's complement, or, say ranging, from 1 to 1024), but there are only 210 = 1024 possible values that can be differentiated in 10 bits.
Erik Eidt
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Thank you, I'm still new to all of "this" and sometimes questions like these can feel overwhelming. – Blind Roach Oct 29 '16 at 17:56