A LevelComputer ScienceCambridge (CIE)Exam Questions4. Processor FundamentalsAssembly Language

Assembly Language (Cambridge (CIE) A Level Computer Science): Exam Questions

Exam code: 9618

1 hour22 questions

5 marks

The following table shows part of the instruction set for a processor. The processor has two registers, the Accumulator (ACC) and the Index Register (IX).

Instruction		Explanation
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
`ADD`	`#n/Bn/&n`	Add the number n to the ACC
`ADD`	`<address>`	Add the contents of the given address to the ACC
`SUB`	`#n/Bn/&n`	Subtract the number n from the ACC
`SUB`	`<address>`	Subtract the contents of the given address 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)
<address> can be an absolute or a symbolic address # denotes a denary number, e.g. #123 B denotes a binary number, e.g. B01001010 & denotes a hexadecimal number, e.g. &4A

The current contents of memory are shown:

Address	Data
`19`	`25`
`20`	`23`
`21`	`2`
`22`	`4`
`23`	`15`
`24`	`50`
`25`	`22`

The current contents of the ACC and IX are shown:

ACC	50
IX	20

Complete the table by writing the content of the ACC and the IX after each set of instructions has run.

	Instructions	ACC content	IX content
1	`LDM #19` `DEC ACC`
2	`LDD 23` `ADD 19`
3	`LDI 25` `INC ACC`
4	`LDR #21` `LDX 2`

How did you do?

Was this exam question helpful?

3 marks

The following table shows part of the instruction set for a processor. The processor has two registers, the Accumulator (ACC) and the Index Register (IX).

Instruction		Explanation
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
`ADD`	`#n/Bn/&n`	Add the number n to the ACC
`ADD`	`<address>`	Add the contents of the given address to the ACC
`SUB`	`#n/Bn/&n`	Subtract the number n from the ACC
`SUB`	`<address>`	Subtract the contents of the given address 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)
<address> can be an absolute or a symbolic address # denotes a denary number, e.g. #123 B denotes a binary number, e.g. B01001010 & denotes a hexadecimal number, e.g. &4A

The current contents of memory are shown:

Address	Data
50	54
51	55
52	50
53	52
54	100
55	25
56	50

The current contents of the ACC and IX are shown:

ACC	50
IX	45

Complete the table by writing the content of the ACC after each program has run.

	Instructions	ACC content
1	`LDD 50` `ADD #4` `ADD 54`
2	`LDI 53` `DEC ACC` `ADD 56`
3	`LDM #55` `SUB #5`

How did you do?

Was this exam question helpful?

4 marks

The following table shows part of the instruction set for a processor. The processor has two registers: the Accumulator (ACC) and an Index Register (IX).

Instruction		Explanation
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
`ADD`	`#n/Bn/&n`	Add the number n to the ACC
`ADD`	`<address>`	Add the contents of the given address to the ACC
`SUB`	`#n/Bn/&n`	Subtract the number n from the ACC
`SUB`	`<address>`	Subtract the contents of the given address from the ACC
`INC`	`<register>`	Add 1 to the contents of the register (ACC or IX)
<address> can be an absolute or a symbolic address # denotes a denary number, e.g. #123 B denotes a binary number, e.g. B01001010 & denotes a hexadecimal number, e.g. &4A

The current contents of memory are shown:

Address	Data
19	24
20	2
21	1
22	3
23	5
24	4
25	22

The current contents of the ACC and IX are shown:

ACC	12
IX	1

Complete the table by writing the content of the ACC after each program has run.

Program number	Instructions	ACC content
1	`LDD 20` `ADD #2`
2	`LDX 22`
3	`LDI 25` `INC ACC` `SUB 22`
4	`LDD 19` `LDM #5` `LDM #25`

How did you do?

Was this exam question helpful?

4 marks

The following table shows part of the instruction set for a processor. The processor has two registers: the Accumulator (ACC) and an Index Register (IX).

Instruction		Explanation
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
`ADD`	`#n/Bn/&n`	Add the number n to the ACC
`ADD`	`<address>`	Add the contents of the given address to the ACC
`SUB`	`#n/Bn/&n`	Subtract the number n from the ACC
`SUB`	`<address>`	Subtract the contents of the given address from the ACC
`INC`	`<register>`	Add 1 to the contents of the register (ACC or IX)
<address> can be an absolute or a symbolic address # denotes a denary number, e.g. #123 B denotes a binary number, e.g. B01001010 & denotes a hexadecimal number, e.g. &4A

The current contents of memory are shown:

Address	Data
10	1
11	3
12	5
13	11
14	10
15	16
16	12

The current contents of the ACC and IX are shown:

ACC	10
IX	0

Complete the table by writing the content of the ACC after each program has run.

Program number	Instructions	ACC content
1	`LDI 15` `SUB #1`
2	`LDD 14` `ADD 11`
3	`LDM #11` `ADD #3` `SUB 16`
4	`LDR #2` `LDX 14` `ADD #2`

How did you do?

Was this exam question helpful?

4 marks

The following table shows part of the instruction set for a processor. The processor has two registers: the Accumulator (ACC) and an Index Register (IX).

Instruction		Explanation
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
`ADD`	`#n/Bn/&n`	Add the number n to the ACC
`ADD`	`<address>`	Add the contents of the given address to the ACC
`SUB`	`#n/Bn/&n`	Subtract the number n from the ACC
`SUB`	`<address>`	Subtract the contents of the given address from the ACC
`INC`	`<register>`	Add 1 to the contents of the register (ACC or IX)
<address> can be an absolute or a symbolic address # denotes a denary number, e.g. #123 B denotes a binary number, e.g. B01001010 & denotes a hexadecimal number, e.g. &4A

The current contents of memory are shown:

Address	Data
48	51
49	6
50	48
51	50
52	49
53	50
54	6

The current contents of the ACC and IX are shown:

ACC	2
IX	50

Complete the table by writing the content of the ACC after each program has run.

Program number	Instructions	ACC content
1	`LDM #50` `INC ACC` `SUB #1`
2	`LDI 51` `ADD 52`
3	`LDR #2` `LDX 50` `DEC ACC`
4	`LDD 52` `SUB 54` `INC ACC`

How did you do?

Was this exam question helpful?

1 mark

Identify the purpose of the first pass of a two-pass assembler.

How did you do?

6 marks

The following table shows part of the instruction set for a processor. The processor has two registers, the Accumulator (ACC) and the Index Register (IX).

Instruction		Explanation
Opcode	Operand	Explanation
`LDR`	`#n`	Immediate addressing. Load the number n to IX
`STO`	`<address>`	Store contents of ACC at the given address
`ADD`	`<address>`	Add the contents of the given address to the ACC
`INC`	`<register>`	Add 1 to the contents of the register (ACC or IX)
`CMP`	`#n`	Compare the contents of ACC with number n
`JPE`	`<address>`	Following a compare instruction, jump to <address> if the compare was True
`OUT`		Output to the screen the character whose ASCII value is stored in ACC
<address> can be an absolute or symbolic address # denotes a denary number, e.g. #123

(i) Give one example of an instruction that belongs to each of the following instruction groups.

Only use the instructions given in the table. Each instruction must have a suitable operand.

Data movement ...........................

Arithmetic operation ...................

Conditional instruction ................

(3)

(ii) The instruction LDR #2 uses immediate addressing.

Give one similarity and one difference between direct addressing and indexed addressing.

(2)

(iii) Identify one other mode of addressing.

(1)

How did you do?

Was this exam question helpful?

4 marks

The following table shows part of the instruction set for a processor. The processor has two registers, the Accumulator (ACC) and the Index Register (IX).

Opcode	Operand	Explanation
Instruction		Explanation
`LDD`	`<address>`	Direct addressing. Load the contents of the location at the given 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
`STO`	`<address>`	Store the contents of ACC at the given address
`ADD`	`#n`	Add the denary number n to the ACC
`JMP`	`<address>`	Jump to the given address
`INC`	`<register>`	Add 1 to the contents of the register (ACC or IX)
`CMP`	`<address>`	Compare the contents of ACC with the contents of <address>
`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
`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
<address> can be an absolute or a symbolic address # denotes a denary number, e.g. #123

The instructions in the processor’s instruction set can be grouped according to their function.

Identify the instruction group for each of the following opcodes.

IN .............................

ADD .........................

JPE ...........................

CMI ...........................

How did you do?

1 mark

The current contents of main memory and selected values from the ASCII character set are given on page 15.

Trace the program currently in memory using the trace table when the input is ‘1’.

Table with computer instructions, memory addresses, and ASCII values. Includes instruction and output columns, memory values for addresses 100-111.

How did you do?

Was this exam question helpful?

2 marks

Assembly language instructions can use different modes of addressing.

Complete the table by writing the name of the addressing mode that each description describes.

Description	Addressing mode
The address of the data is formed by adding the contents of the Index Register to the operand.
The operand gives an offset. This offset is added to a base address to form the address of the data, so the program still works if it is loaded at a different place in memory.

How did you do?

Was this exam question helpful?

3 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (IX).

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

<address> can be an absolute or a symbolic address. # denotes a denary number, for example #123.

The current contents of main memory are:

Address	Data
100	104
101	55
102	100
103	12
104	78
105	101

The Index Register (IX) holds the value 2.

Write the single instruction that loads the value 78 into the ACC using each of the addressing modes shown in the table below.

Addressing mode	Instruction
Direct
Indirect
Indexed

How did you do?

Was this exam question helpful?

4 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (IX).

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
`ADD`	`<address>`	Add the contents of the given address to the ACC

<address> can be an absolute or a symbolic address. # denotes a denary number, for example #123.

The current contents of main memory are:

Address	Data
200	204
201	30
202	206
203	201
204	45
205	202
206	18
207	200
208	75

Each program starts with ACC holding 0 and IX holding 3.

Complete the table by writing the contents of the ACC after each program has run.

Program number	Code	ACC contents
1	`LDX 201`
2	`LDI 205`
3	`LDR #1` `LDX 205` `ADD 201`
4	`LDM #207` `LDI 207`

How did you do?

Was this exam question helpful?

5 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC).

Instruction		Explanation
Opcode	Operand	Explanation
`LDD`	`<address>`	Direct addressing. Load the contents of the location at the given address to ACC
`ADD`	`#n`	Add the denary number n to the ACC
`STO`	`<address>`	Store the contents of ACC at the given address
`END`		Return control to the operating system
`<address>` can be an absolute or a symbolic address `#` denotes a denary number, e.g. `#123`

A programmer writes this assembly language program.

LDD 300

ADD #4

STO 301

END

(i) The program is translated so that the processor can run it. State how many instructions the translated program contains.

(ii) Explain your answer to part (i).

(iii) In its translated form, ADD #4 is one pattern of bits made up of an opcode and an operand. Describe what the processor does with each of these two parts.

How did you do?

Was this exam question helpful?

2 marks

The instructions in a processor's instruction set are placed in groups. Each group contains instructions that carry out a similar kind of task.

The table shows four instructions from one processor's instruction set.

Instruction		Explanation
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
`ADD`	`#n`	Add the denary number n to the ACC
`SUB`	`#n`	Subtract the denary number n from the ACC

(i) Name the instruction group that IN and OUT belong to.

(ii) Name the instruction group that ADD and SUB belong to.

How did you do?

Was this exam question helpful?

3 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (IX).

Instruction		Explanation
Opcode	Operand	Explanation
`LDD`	`<address>`	Direct addressing. Load the contents of the location at the given address to ACC
`STO`	`<address>`	Store the contents of ACC at the given address
`ADD`	`<address>`	Add the contents of the given address to the ACC
`INC`	`<register>`	Add 1 to the contents of the register (ACC or IX)
`CMP`	`#n`	Compare the contents of ACC with number n
`JMP`	`<address>`	Jump to the given address
`JPE`	`<address>`	Following a compare instruction, jump to the given address if the compare was True
`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
`<address>` can be an absolute or a symbolic address `#` denotes a denary number, e.g. `#123`

LDD and STO both belong to the data movement group.

Name one other instruction group, describe what the instructions in that group do, and give one example instruction from the table that belongs to it. Your example must include a suitable operand where the instruction needs one.

How did you do?

Was this exam question helpful?

3 marks

A cinema has 250 seats. The memory location at address 300 holds the number of seats that are still free.

A program written in assembly language reads that number and outputs one character to describe the cinema: F if no seats are free, E if every seat is free, and S if some seats are free.

The program is stored in main memory starting at address 50. Each instruction is stored in one memory location.

Address	Label	Instruction
50		`LDD 300`
51		`CMP #0`
52		`JPE FULL`
53		`CMP #250`
54		`JPE EMPTY`
55		`JMP SOME`
56	FULL	`LDM #70`
57		`OUT`
58		`END`
59	EMPTY	`LDM #69`
60		`OUT`
61		`END`
62	SOME	`LDM #83`
63		`OUT`
64		`END`

A two-pass assembler translates this program. During the first pass the assembler builds a symbol table.

Complete the symbol table.

Label	Address

How did you do?

Was this exam question helpful?

4 marks

A program written in assembly language divides the number stored at address 200 by the number stored at address 201, using repeated subtraction. It stores the result at address 202. The value at address 200 is an exact multiple of the value at address 201, and address 202 holds 0 before the program starts.

The program is stored in main memory starting at address 40. Each instruction is stored in one memory location.

Address	Label	Instruction
40		`LDD 200`
41	AGAIN	`CMP #0`
42		`JPE DONE`
43		`SUB 201`
44		`STO 200`
45		`LDD 202`
46		`INC ACC`
47		`STO 202`
48		`LDD 200`
49		`JMP AGAIN`
50	DONE	`END`

The program is translated by a two-pass assembler.

Explain why this program cannot be translated into object code in a single pass.

How did you do?

Was this exam question helpful?

5 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC).

Instruction		Explanation
Opcode	Operand	Explanation
`LDD`	`<address>`	Direct addressing. Load the contents of the location at the given address to ACC
`ADD`	`#n`	Add the denary number n to the ACC
`STO`	`<address>`	Store the contents of ACC at the given address
`CMP`	`#n`	Compare the contents of ACC with number n
`JPN`	`<address>`	Following a compare instruction, jump to the given address if the compare was False
`JMP`	`<address>`	Jump to the given address
`END`		Return control to the operating system
`<address>` can be an absolute or a symbolic address `#` denotes a denary number, e.g. `#123`

A program is stored in memory at addresses 100 to 105. Address 200 holds the value 3 before the program starts.

100   LDD 200

101   ADD #1

102   STO 200

103   CMP #10

104   JPN 100

105   END

(i) Explain why the instruction JPN 100 at address 104 cannot do its job unless the instruction at address 103 is also present.

(ii) The instruction at address 104 is replaced by JMP 100. Describe the effect this change has when the program runs, and name the instruction group that JMP belongs to.

How did you do?

Was this exam question helpful?

2 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC).

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
`ADD`	`#n`	Add the number n to the ACC
`SUB`	`address`	Subtract the contents of the given address from the ACC

address can be an absolute or a symbolic address.

# denotes a denary number, for example #123.

The current contents of memory are shown.

Address	Data
60	12
61	9
62	7
63	25

The current content of the ACC is 4. The value 4 is reloaded into the ACC before each program is run.

Complete the table by writing the content of the ACC after each program has run.

Program number	Code	ACC content
1	`LDD 62` `ADD #5`
2	`LDM #62` `SUB 61`

How did you do?

Was this exam question helpful?

3 marks

The table shows part of the instruction set for a processor. The processor has two registers, the Accumulator (ACC) and the Index Register (IX).

Opcode	Operand	Explanation
`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 the given 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
`DEC`	`register`	Subtract 1 from the contents of the register (ACC or IX)

address can be an absolute or a symbolic address.

# denotes a denary number, for example #123.

The current contents of memory are shown.

Address	Data
70	74
71	6
72	19
73	71
74	33
75	8
76	41

The current contents of the ACC and the IX are shown.

Register	Content
ACC	5
IX	3

The ACC and the IX are reset to these values before each set of instructions is run.

Complete the table by writing the content of the ACC and the content of the IX after each set of instructions has run.

Set number	Instructions	ACC content	IX content
1	`LDR #2` `LDX 73`
2	`LDI 73`
3	`LDD 72` `DEC ACC`

How did you do?

Was this exam question helpful?

5 marks

The table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (IX).

Opcode	Operand	Explanation
`LDR`	`#n`	Immediate addressing. Load the number n to IX
`LDX`	`address`	Indexed addressing. Form the address from the given address + the contents of the Index Register. Copy the contents of this calculated address to ACC
`CMP`	`#n`	Compare the contents of ACC with number n
`JPE`	`address`	Following a compare instruction, jump to the given address if the compare was True
`JMP`	`address`	Jump to the given address
`INC`	`register`	Add 1 to the contents of the register (ACC or IX)
`OUT`		Output to the screen the character whose ASCII value is stored in ACC
`END`		Return control to the operating system

address can be an absolute or a symbolic address.

# denotes a denary number, for example #123.

The current contents of memory are shown. The value 0 at address 153 marks the end of the message.

Address	Data
150	72
151	73
152	33
153	0

The ASCII values used by the program are shown.

ASCII value	Character
72	H
73	I
33	!

The program is shown.

Label	Opcode	Operand
	`LDR`	`#0`
LOOP	`LDX`	`150`
	`CMP`	`#0`
	`JPE`	`FINISH`
	`OUT`
	`INC`	`IX`
	`JMP`	`LOOP`
FINISH	`END`

The trace table records what happens on each pass through the loop. The first row has been completed for you.

Complete the trace table.

Pass through the loop	Content of the ACC after `LDX 150`	Character output during this pass	Content of the IX at the end of this pass
1	72	H	1
2
3
4

How did you do?

Was this exam question helpful?