This problem considers instruction set architectures for three architecture styl

Question

This problem considers instruction set architectures for three architecture styl accumulator architecture, a stack architecture, and a two-address load-store architecture. (You can find more information about these architectures in Appendix A, Section A.2 of the text. Note that the MIPS is a three-address load-store architecture, yet another style.) Make the following assumptions about all three instruction set architectures: 3. es, an . The opcode is always one byte. . All memory addresses are two bytes All data operands are two bytes. Each architecture uses a fixed length instruction set is fixed length (every instruction used by the architecture has the same integer number of bytes). . An accumulator architecture is simple, but still used on many microcontrollers. They accumulator as one of its two inputs, and puts its result back into the accumulator. have a special working register called the accumulator; every math operation uses the Suppose that the accumulator architecture has the following assembly language instructions: LOAD

Explanation / Answer

Answer for 3.a).

How many bytes in an instruction are needed for the accumulator architecture?

Accumulator Architecture:

     An Accumulator architecture is based on the concept that one register (called accumulator) will store one operand on which calculation is to be performed and that register itself will store the result after the manipulation operation is performed and the second or other operands will be stored in some other registers or some other memory locations.

A stack Architecture:

               

                Operands are on the top of the stack and the result of an operation is also stored on the top of the stack.

A two address load-store architecture:

                                Operands are stored in registers or specific memory locations and results are also stored in register/memory.

Below is the instruction format for accumulator architecture:

As it is given that opcode is of 1 byte i.e., 8-bits and memory addresses are of 2 bytes i.e., 16-bits.

For accumulator architecture, there will be two fields opcode and Address of memory where an operand is stored which will be loaded into accumulator so no need of separate field for an accumulator.

Now suppose we need to find the bytes required for :

Load

Store

Add

So here instruction count is =3

Now total bits required for instructions = number of bits for opcode + number of bits for address

= (3*8)+(3*16)

= 24 + 48

= 72 bits

= 72/8 bytes

= 9 bytes

Ans. 3.b).

Assembly language code for C=A+B+C+D on the accumulator architecture:

Load A

Add B

Add C

Add D

Store C

OPCODE ADDRESS OF OPERAND 8-bits 16 bits

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