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Option 1:Parallelism is high in the horizontal microprogrammed control unit as compared to a vertical microprogrammed control unit. True, Parallelism is high in horizontal microprogramming as several operations on different registers can be performed simultaneously.
Option 2: Hardwired control unit is slower compared to the microprogrammed control unit. False, a Hardwired control unit is faster as compared to the microprogrammed control unit as there won’t be a delay of fetch, decoding, and executing the control instructions in the case of the hardwired control unit.
Option 3: In 2’s complement sum carry flag and overflow are the same. False, In unsigned numbers, carry out is equivalent to overflow. But in two's complement, carry out tells you nothing about overflow.
Option 4: In 2’s complement sum if the sum of two negative numbers yields a positive result, the sum has overflowed. True, the Following are the rules for detecting overflow in a two's complement sum:
If the sum of two positive numbers yields a negative result, the sum has overflowed.
If the sum of two negative numbers yields a positive result, the sum has overflowed.
Otherwise, the sum has not overflowed.
Hence the correct answer is option 1 and option 4.
Commercial computers frequently have this. Two addresses can be supplied in the instruction in this case. Instead of the result being saved in the accumulator as it was in prior address instructions, the result can now be stored in many locations, however, this requires extra bits to indicate the address.
The given data,
The CPU supports op-code size = 16 bits
Address size = 4 bits
We have two operands so it requires the 2 x 4 bits =8
And remaining 16-8 bits can be used for two address instructions.
i.e 8 bits.
Maximum number of instructions = 28 = 256
Hence the correct answer is 256.
RISC is implemented using hardwire control unit. RISC uses registers instead of memory. Registers are small in size and are on the same chip on which ALU and control unit are present. RISC architecture is shown below.
A processor has 300 distinct instructions and 70 general-purpose registers. A 32-bit instruction word has an opcode, two register operands, and an immediate operand. The number of bits available for the immediate operand field is_____
Distinct instructions= 300
General-purpose registers = 70
opcode instruction word = 32-bit
Two register operands and an immediate operand.
Each instruction has 32 bits. To support 300 instructions, the opcode must contain 9-bits.
Register operand1 requires 7 bits, since the total registers are 70, Register operand 2 also requires 7 bits.
So, 32 - (9+7+7) = 9 bits are left over for immediate operand.
Option 1: In the immediate addressing mode the operand is placed in the instruction itself.
True, An immediate mode instruction has an operand field rather than an address field. The operand field contains the actual operand to be used in conjunction with the operation specified in the instruction.
Option 2: One-byte machine instruction consists of the only operand. False, The machine instructions which consist of the only opcode are called one-byte machine instructions.
Option 3: Indirect addressing mode is suitable for implementing pointers in C. True, Indirect addressing mode is suitable for implementing pointers in C. The instruction includes the address of the place where the target address is stored in this style of addressing. As a result, it is indirectly storing the target site's address in another memory location.
Option 4: Displacement addressing mode is similar to the register indirect addressing mode.
True, The displacement addressing mode is equal to the register indirect addressing mode, with the exception that the effective address of the operand is formed by adding an offset (or displacement) maintained in the instruction to the contents of the register provided in the instruction.
Hence the correct answer is option 1, option 3 and option 4.
All the operations will be performed in the Accumulator register(AC).
The load operation is used to fetch the value from register or memory to accumulator.
The store operation is used to store the value from the accumulator to register or memory.
The one address instructions for the given equations are:
Hence, the correct answer is "option 3".
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