Assembly Language

Opcode	Operand	Explanation
LDM	#n	Immediate addressing. Load the number n to ACC
LDD	<address>	Direct addressing. Load the contents of the location at the given address to ACC
LDI	<address>	Indirect addressing. The address to be used is at the given address. Load the contents of this second address to ACC
LDX	<address>	Indexed addressing. Form the address from <address> + the contents of the index register. Copy the contents of this calculated address to ACC
LDR	n	Immediate addressing. Load the number n to IX
MOV	<register>	Move the contents of the accumulator to the given register (IX)
STO	<address>	Store the contents of ACC at the given address
ADD	<address>	Add the contents of the given address to the ACC
ADD	#n/Bn/&n	Add the number n to the ACC
SUB	<address>	Subtract the contents of the given address from the ACC
SUB	#n/Bn/&n	Subtract the number n from the ACC
INC	<register>	Add 1 to the contents of the register (ACC or IX)
DEC	<register>	Subtract 1 from the contents of the register (ACC or IX)
JMP	<address>	Jump to the given address
CMP	<address>	Compare the contents of ACC with the contents of <address>
CMP	#n	Compare the contents of ACC with number n
CMI	<address>	Indirect addressing. The address to be used is at the given address. Compare the contents of ACC with the contents of this second address
JPE	<address>	Following a compare instruction, jump to <address> if the compare was True
JPN	<address>	Following a compare instruction, jump to <address> if the compare was False
IN		Key in a character and store its ASCII value in ACC
OUT		Output to the screen the character whose ASCII value is stored in ACC
END		Return control to the operating system

提示

All questions will assume there is only one general purpose register available (Accumulator) ACC denotes Accumulator
IX denotes Index Register
<address> can be an absolute or symbolic address
# denotes a denary number, e.g. #123
B denotes a binary number, e.g. B01001010 & denotes a hexadecimal number, e.g. &4A

Assembly language and machine code

The only programming language that a CPU can use is machine code.
Every different type of computer/chip has its own set of machine code instructions.
A computer program stored in main memory is a series of machine code instructions that the CPU can automatically carry out during the fetch-execute cycle.
The first programming language to be developed was assembly language, this is closely related to machine code and uses mnemonics instead of binary.

Stages of assembly

Before a program written in assembly language (source code) can be executed, it needs to be translated into machine code.
The translation is performed by a program called an assembler.
An assembler translates each assembly language instruction into a machine code instruction.
An assembler also checks the syntax of the assembly language program to ensure that only opcodes from the appropriate machine code instruction set are used.

A two pass assembler produces an object program in machine code that can be stored, loaded then executed at a later stage.
This requires the use of another program called a loader.
Two pass assemblers need to scan the source program twice, so they can replace labels in the assembly program with memory addresses in the machine code program.

Pass 1

Read the assembly language program one line at a time.
Ignore anything not required, such as comments.
Allocate a memory address for the line of code.
Check the opcode is in the instruction set.
Add any new labels to the symbol table with the address, if known.
Place address of labelled instruction in the symbol table.

Pass 2

Read the assembly language program one line at a time.
Generate object code, including opcode and operand, from the symbol table

generated in Pass 1.

Save or execute the program.

The only programming language that a CPU can use is .

machine code

[0/1]

The translates each assembly language instruction into a machine code instruction.

[0/1]

Assembly language instructions

Data movement instructions

Opcode	Operand	Explanation
LDM	#n	Immediate addressing. Load the number n to ACC
LDD	<address>	Direct addressing. Load the contents of the location at the given address to ACC
LDI	<address>	Indirect addressing. The address to be used is at the given address. Load the contents of this second address to ACC
LDX	<address>	Indexed addressing. Form the address from <address> + the contents of the index register. Copy the contents of this calculated address to ACC
LDR	n	Immediate addressing. Load the number n to IX
MOV	<register>	Move the contents of the accumulator to the given register
STO	<address>	Store the contents of ACC at the given address

Input and output of data instructions

Opcode	Operand	Explanation
IN		Key in a character and store its ASCII value in ACC
OUT		Output to the screen the character whose ASCII value is stored in ACC

No opcode is required as a single character is either input to the accumulator or output

Arithmetic operation instructions

Opcode	Operand	Explanation
ADD	<address>	Add the contents of the given address to the ACC
ADD	#n/Bn/&n	Add the number n to the ACC
SUB	<address>	Subtract the contents of the given address from the ACC
SUB	#n/Bn/&n	Subtract the number n from the ACC
INC	<register>	Add 1 to the contents of the register (ACC or IX)
DEC	<register>	Subtract 1 from the contents of the register (ACC or IX)

Answers to calculations are always stored in the accumulator

Unconditional and conditional instructions

Opcode	Operand	Explanation
JMP	<address>	Jump to the given address
JPE	<address>	Following a compare instruction, jump to <address> if the compare was True
JPN	<address>	Following a compare instruction, jump to <address> if the compare was False
END		Return control to the operating system

Jump means change the PC to the address specified, so the next instruction to beexecuted is the one stored at the specified address, not the one stored at the next location in memory

Compare instructions

Opcode	Operand	Explanation
CMP	<address>	Compare the contents of ACC with the contents of <address>
CMP	#n	Compare the contents of ACC with number n
CMI	<address>	Indirect addressing. The address to be used is at the given address. Compare the contents of ACC with the contents of this second address

The contents of the accumulator are always compared

Addressing modes

Assembly language and machine code programs use different addressing modes depending on the requirements of the program.

Absolute addressing – the contents of the memory location in the operand are used. For example, LDD 200.
Direct addressing – the contents of the memory location in the operand are used. For example, LDD 200.
Indirect addressing – the contents of the contents of the memory location in the operand are used. For example, LDI 200.
Indexed addressing – the contents of the memory location found by adding the contents of the index register (IR) to the address of the memory location in the operand are used. For example, LDX 200.
Immediate addressing – the value of the operand only is used. For example, LDM #200.
Relative addressing – the memory address used is the current memory address added to the operand. For example, JMR #5.
Symbolic addressing – only used in assembly language programming. A label is used instead of a value. For example, LDD MyStore.

Simple assembly language programs

A program written in assembly language will need many more instructions than a program written in a high-level language to perform the same task.

In a high-level language, adding a list of numbers together and storing the answer would typically be written using a loop.

CIR	Opcode	Operand	ACC	IX	Counter 118	Total 119
100	LDM	#0	0
101	STO	total	0			0
102	STO	counter	0		0	0
103	LDR	#0	0	0	0	0
104	LDX	number	5	0	0	0
105	ADD	total	5	0	0	0
106	STO	total	5	0	0	5
107	INC	IX	5	1	0	5
108	LDD	counter	0	1	0	5
109	INC	ACC	1	1	0	5
110	STO	counter	1	1	1	5
111	CMP	#3	1	1	1	5
112	JPN	loop	1	1	1	5
104	LDX	number	7	1	1	5
105	ADD	total	12	1	1	5
106	STO	total	12	1	1	12
107	INC	IX	12	2	1	12
108	LDD	counter	1	2	1	12
109	INC	ACC	2	2	1	12
110	STO	counter	2	2	2	12
111	CMP	#3	2	2	2	12
112	JPN	loop	2	2	2	12
104	LDX	number	3	2	2	12
105	ADD	total	15	2	2	12
106	STO	total	15	2	2	15
107	INC	IX	15	3	2	15
108	LDD	counter	2	3	2	15
109	INC	ACC	3	3	2	15
110	STO	counter	3	3	3	15
111	CMP	#3	3	3	3	15
112	JPN	loop	3	3	3	15
113	END

State the contents of the accumulator after the following instructions have been executed. The memory location with address 200 contains 300, the memory location with address 300 contains 50.

LDM #200 (ACC: )
LDD 200 (ACC: )
LDI 200 (ACC: )

[0/1]

Write an assembly language instruction to:

compare the accumulator with 5:
jump to address 100 if the comparison is true:

[0/2]

What is the value of total in the end?

Label	Opcode	Operand
	LDD	number1
	SUB	number2
	ADD	number3
	CMP	#10
	JPE	nomore
	ADD	number4
nomore:	STO	total
	END

number1:	#30
number2:	#40
number3:	#20
number4:	#50
total:	#0

[0/3]