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A machine has a 32-bit architecture, with 1-word long instructions. It has 64 registers, each of which is 32 bits long. It needs to support 45 instructions, which have an immediate operand in addition to two register operands. Assuming that the immediate operand is an unsigned integer, the maximum value of the immediate operand is ______ .
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- 183
- 1683
- 16383
- 10381
Correct Option: C
(i) Given is 32-bit architecture machine with 1 word long instruction; so, the instruction size is 32 bit.
(ii) As the machine need to support 45 instructions, the number of bits required in the “opcode field” is 6 bits, because,
5-bits: 25 = 32 (Not sufficient)
6-bits: 26 = 64 (Sufficient)
(iii) As the machine has 64 registers, so to uniquely identify a register out of those 64 registers, we need 6-bits. (using the similar argument as above) As each instruction has 2 register operands, so 6 + 6 = 12 bits are needed for them. So, finally instruction will took like as follows:
(Total = 32 bits)
(iv) As 14-bits are available for immediate operand, the max value will be 214 = 16384. However one value will be used for “Zero” also as the operand is an “unsigned integer”. So, finally, the maximum value that the immediate operand can take = 214 – 1 = 16383.
