Questions about the operation of this C68MX11 CPU

[Michael Neo]

Homework Statement

This question refers to the C68MX11 CPU, details of which accompany this assignment.
(Data sheet: http://www.moxsyn.com/data_sheets/C68MX11_cpu.pdf)

a) With reference to the block diagram of the CPU, what is the purpose of the program stack and the accumulators A, B and D?

b) Again with reference to the block diagram, outline how the processor would add two numbers held in ACC.A and ACC.B, the result being stored in memory.
Identify the addressing modes that would be employed in this addition.

c) Give the address range in hex that the PC could access.

Relevant Equations

None.

Part (a)

The Purpose of the Accumulators

The accumulator is a register for short-term, intermediate storage of arithmetic and logic data in a CPU.

(The term "accumulator" is rarely used in reference to contemporary CPUs, having been replaced around the turn of the millennium by the term "register". In a modern computers, any register can function as an accumulator. )

With reference to the block diagram of the CPU, although the data busses have a width of 8-bits most of the instructions have 16-bit equivalent instructions.

The C68MX11 offers multiply, add, subtract, compare, increment and decrement, load & store, and shift instructions of 16-bit operands.

The CPU consists of 8-bit accumulators used to hold operands and results of arithmetic calculations or data manipulations.

The accumulator A and B can be combined into a 16-bit double accumulator D that can store these 16-bit equivalent instructions.

From the data sheet:

The Purpose of the Program Stack

From the data sheet: “The CPU automatically supports a program stack. This stack may be located anywhere in the 64 Kbyte address space through the stack pointer and may be any size up to the amount of memory available in the system.”

Definition
A stack is a data structure.
A stack is a linear data structure which follows a particular order in which the operations are performed.
A stack is an array of memory dedicated to temporary storage.

Structure
A stack is, typically, a LIFO data structure.
A stack stores data from the top down.

Function
A stack pointer is a register that stores, and is used to indicate, the address of the last item put onto the stack.

• When a new data item is entered or "pushed" onto the top of a stack, the stack pointer increments to the next physical memory address, and the new item is copied to that address.
• When a data item is "pulled" from the top of a stack, the item is copied from the address of the stack pointer, and the stack pointer decrements to the next available item at the top of the stack.

Summary:

• Stack is a linear data structure.
• Typically, but not exclusively, LIFO.
• Stack Pointer is a register that indicates address of last item put on stack.
• Stack Pointer decreases when you put an item on the stack.
• Stack Pointer increases when you pull an item from the stack.

In this CPU, the stack is located anywhere in the 64 Kbyte address space via the stack pointer, and can be up to the amount of memory available in the system.

Part (b)

ACC.A and ACC.B are accumulators.

Accumulator A is the most significant octet while Accumulator B is the least significant octet.

The contents of these two 8-bit accumulators are fed into the ALU and then via the CCR are stored in Accumulator D.

Accumulators are used to store intermediate arithmetic and logic results and then fed to the next immediate level, i.e. the ALU.

OR

Add values stored in ACC.A and ACC.B and put the result in ACC.D, then run a separate cycle to store ACC.D in memory via the address bus and data port.

The binary operands are brought to the ALU via two inputs from ACC.A and ACC.B via the bus.

The addition takes place in the 16-bit ALU by adder circuits.

The 16 bit index registers IX and IY are used for indexed addressing modes.

The addressing modes that would be employed in this addition are:

1. Direct
2. Extended
3. Indexed

Part (c)

1 KB = 1024 Bytes
64 KB = 65536 Bytes
So, address range = 0 to 65535

Convert decimal notation to hexadecimal,
0 = 0x0
65535 = 0xFFFF

Address range in hex,
0x0 to 0xFFFF

 

[Tom.G]

Part (a)
a) With reference to the block diagram of the CPU, what is the purpose of the program stack

You have described how the stack operates, but the problem asks why there is one.

Part (b) outline how the processor would add two numbers

The "OR in the middle of your answer(s) is problematic. Both parts are accurate. The first part defines the registers and their general usage, which is useful. But by itself is an incomplete response to the problem statement.

Part (c) Address range
I agree.

 

[Michael Neo]

Part (a)

Program stacks are used to support: expression evaluation, subroutine return address storage, dynamically allocated local variable storage, and subroutine parameter passing.

Expression Evalution

To see why a program stacks is used, consider how the following arithmetic expression would be computed:
X = (A + B) * (C + D)
First, A and B would be added together.
Then, this intermediate result must be saved somewhere.
It is pushed onto the expression evaluation stack.
Next, C and D are added, and the result is also pushed onto the expression evaluation stack.
Finally, the top two stack elements (A+B and C+D) are multiplied and the result is stored in X.
The expression evaluation stack provides automatic management of intermediate results of expressions, and allows as many levels of precedence in the expression as there are available stack elements.

Subroutine Return Address Storage

...

Dynamically Allocated Local Variable Storage

...

Subroutine Parameter Passing

...

Are general reasons for having a stack applicable to this CPU. Is that all that is required?
It is a relatively old CPU.

Part (b)

ACC.A and ACC.B are accumulators: accumulator A and accumulator B respectively.
Accumulator A is the most significant octet while Accumulator B is the least significant octet.
These two general-purpose 8-bit accumulators are used to store operands and results of arithmetic calculations or data manipulations and then fed to the next immediate level, i.e. the contents of these two 8-bit accumulators are fed into the 16-bit ALU.
The binary operands are brought to the 16-bit ALU via two inputs from ACC.A and ACC.B via the bus.
The addition takes place in the 16-bit ALU by adder circuits.
Then a separate cycle is run to store ACC.D (accumulator D) in memory via the address bus and data port, via the CCR, are stored in 16-bit Accumulator D.

The 16 bit index registers IX and IY are used for indexed addressing modes.
The addressing modes that would be employed in this addition are:

1. Direct
2. Extended
3. Indexed

This doesn't seem correct - A and B to ALU via CCR to D to CCR to memory? Or A and B (which comprise D) to ALU via CCR to memory?

 

[Tom.G]

Are general reasons for having a stack applicable to this CPU.

Can't tell without a detailed description of the instruction set.
However the problem statement was:

what is the purpose of the program stack

This doesn't seem correct - A and B to ALU via CCR to D to CCR to memory? Or A and B (which comprise D) to ALU via CCR to memory?

Yea, it does seem a bit off.

• A and B to ALU via CCR to D
  and
  Then a separate cycle is run to store ACC.D (accumulator D) in memory via the address bus and data port, via the CCR, are stored in 16-bit Accumulator D. {a leftover sentence fragment here?}
  
  I don't see where A & B go thru the CCR to get to the ALU, it seems to be a direct route; {register}-->ALU. The main purpose of CCR is to hold flag bits that reflect ALU results, so things like the Carry out of an operation can be used as input to the next ALU instruction, or to test if a Conditional Branch instruction should be taken or not.
• Or A and B (which comprise D)
  
  This wording seems to imply that D is a way of treating the A and B registers as a single item. I reality, D is another register that is separate from A and B.

 

[Michael Neo]

This wording seems to imply that D isa way of treating the A and B registsrs as a single item. In reality, D is another register that is separate from A and B.

The confusion stems from the wording of the data sheet:

The CPU consists of two general-purpose 8-bit accumulators used to hold operands and results of arithmetic calculations or data manipulations. The accumulator A and B can be combined into a 16 bit double accumulator D.

So, this actually means the contents of A and B can be placed into D?

 

[Michael Neo]

With reference to the block diagram of the CPU, what is the purpose of the program stack and the accumulators A, B and D?

The program stack
In the data sheet, it states:

“The CPU automatically supports a program stack. This stack may be located anywhere in the 64 Kbyte address space through the stack pointer and may be any size up to the amount of memory available in the system.”

The program stack is a data structure that operates, typically, LIFO.

A stack pointer is a register that stores, and is used to indicate, the address of the last item put onto the stack.

When a new data item is entered or “pushed” onto the top of a stack, the stack pointer increments to the next physical memory address, and the new item is copied to that address.

When a data item is “pulled” from the top of a stack, the time is copied from the address of the stack pointer, and the stack pointer decrements to the next available item at the top of the stack.

The purpose of the program stack is to store information about the active subroutines of a computer program, expression evaluation, subroutine return address storage, dynamically allocated variable storage, and subroutine parameter passing.

[This doesn't seem enough. "With reference to the block diagram of the CPU..." How can the block diagram be specifically related to the purpose of the program stack? I've described, why a program stack is needed and I've described why a stack pointer (special register) is needed to enable this.]

The accumulators A, B and D
In the data sheet, it states:

“Although the data busses have a width of 8 bit most of the instructions have 16 bit equivalent instructions. The C68MX11 offers multiply, add, subtract, compare, increment & decrement, load & store, and shift instructions of 16 bit operands. The CPU consists of two general-purpose 8-bit accumulators used to hold operands and results of arithmetic calculations or data manipulations. The accumulator A and B can be combined into a 16 bit double accumulator D.”

An accumulator is a register for short-term storage of operands, and results of arithmetic calculations or data manipulations.

Accumulator A and B are 8-bit while accumulator D is 16-bit.

However, most of the instructions have 16-bit equivalent instructions but the data busses have a width of 8-bits.

The contents of the 8-bit accumulators A and B can, after manipulation, be stored in 16-bit double accumulator D; after, for example, information from A and B is taken to the 16-bit ALU.

[Again, this doesn't seem enough. It is simply regurgitating the quote from the data sheet. With reference to the block diagram of the CPU..." How can the block diagram be specifically related to acumulators A, B and D?]

 

[Michael Neo]

Part (b)

So, the CCR is not involved at all!

How is the output of the ALU directed to ACC.D?

 

[Tom.G]

The confusion stems from the wording of the data sheet:

The confusion was mine. I misread the data sheet block diagram. The A and B can combine to form the D register. Sorry.

So, the CCR is not involved at all!

Yup, it just holds the status of the most recent ALU operation.

How is the output of the ALU directed to ACC.D?

If there are 16-bit arithmatic instructions, the ALU result must go to both A and B registers. Then 16-bit Store insructions could write the D register (A and B combined) to memory.

I can see you have read that datasheet in more detail than I have! I have less experience with the Motorola 6800 family than I do with the Intel 8080 family of CPUs... and it shows!

I think you have pretty well got it now. Hang in there.

Cheers,
Tom