A 16-bit ISA has the following format: <OPCODE><DR><SR0><SR1> where DR is the Destination register and SR0 and SR1 are source registers. If the ISA defines 16 registers then the maximum number of opcodes support by this format is ________.
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ISA defines 16 registers which means each register is 4-bit in size ( since, 2^4 = 16)
<OPCODE> <DR> <SR0> <SR1> = 4-bit + 4-bit + 4-bit + 4-bit = 16-bit ISA
OPCODE = 4-bit
Hence, maximum number of opcodes supported by this format is ,2^Iopcode sizeI =2^4 = 16
So, for example:
Binary 0101 000 1101 0100
0101 = OPCODE = 4 bits = 2^4
0000 = DR = 4 bits = 2^4
1101 = SR0 = 4 bits = 2^4
0100 = SR1 = 4 bits = 2^4
and the maximum number of opcodes is 4 bits ?
I though in binary in 0101 is 2^3 at the far end left because
I really confuse. and my 2nd though is:
Is it talking about how many possible registers can it takes in how many bits? like:
Am I right about the 2nd though?
So the answer is 4 bits?
In the question it was mentioned that there are 16 registers which means each register can be represented by 4 - bits.
and ISA is <OPCODE> <DR> <SR0> <SR1> which is16bits
<OPCODE> + 4 + 4 + 4 = 16
<OPCODE> = 16 -12 = 4 bits
OPCODE is represented by 4-bits
So, maximum number of opcodes supported by this format is nothing but number of possibilities with four bits which is 2^4 = 16
As u mentioned, 0101 is one representation in binary. You can have 16 such representations in binary with four bits
I'm still not quite sure the answer.
Is it 4 bits OR 16 possibilities?
The answer is 16 number of opcodes
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