What is the maximum number of bits that can be used to represent the IMM value?
IMPORTANT NOTE: If imm is greater than 16 bits, the assembler will truncate the more significant bit positions beyond the sixteenth place. This means the maximum immediate value is 65535.
How many number of bits are there for opcode field if the instruction is in one memory word?
32-bit
A processor has 40 distinct instructions and 24 general purpose registers. A 32-bit instruction word has an opcode, two register operands and am immediate operand. The number of bits available for the immediate operand field is ______….
| Opcode | Register operand | Immediate operand |
|---|---|---|
| 6 bits | 10 bits | x bits |
What is the minimum number of bits required to write a memory type instruction?
The computer allows only 2-address instructions, where one operand can be a register and another can be a memory location. The memory is byte addressable with 64KB (Kilo bytes) in size. The minimum number of bits to encode the instruction will be 28 .
How many bits are needed for the opcode how many bits are left for the address part of the instruction?
An instruction is stored in a word with 24 bits. So, there will be (24-8) = 16 bits for an address part in an instruction. The largest unsigned binary number that can fit into one word of the memory is, (111111111111111111111111)2….Answers:
| Tag | Set | Offset |
|---|---|---|
| 4 | 2 | 2 |
How many R type instructions are possible in a 32-bit architecture?
three different
All instructions in the MIPS R2000 Architecture are 32 bits in length. There are three different instruction formats: R-Type instructions, I-Type instructions, and J-Type instructions….J-Type Instruction Format (Jump):
| 31-26 | 25-0 |
|---|---|
| opcode | target |
How many bits does a machine instruction have?
6.4. A program written in machine language is a series of 32-bit numbers representing the instructions. Like other binary numbers, these instructions can be stored in memory.
How many bits are needed to specify 32 distinct operations?
As the processor has 45 instructions, number of bits for opcode = 6 (2^6 = 64) Total bits occupied by 2 registers and opcode = 6 + 6 + 6 =18. As instruction size given is 32 bits, remaining bit left for immediate operand = 32-18 = 14 bits.
What is a 32-bit operating system?
A 32-bit system can access 232 different memory addresses, i.e 4 GB of RAM or physical memory ideally, it can access more than 4 GB of RAM also. A 64-bit system can access 264 different memory addresses, i.e actually 18-Quintillion bytes of RAM.