This document provides information about assembly language programming basics for the 8085 microprocessor. It begins with definitions of assembly language and its advantages over machine language. It then describes the basic features of a PC's hardware including registers, memory sizes and data types. The next sections cover binary and hexadecimal number systems and binary arithmetic operations. The remaining sections provide details about the different instruction sets of the 8085 including data transfer, logical, branching and arithmetic instructions as well as the various addressing modes.

1. Course :- Microprocessor
Unit :- Programming with 8085
Topic:- Assembly Language Programming Basics
Prepared By
Mr Kupade K A
Assist Prof ETC Dept.
SITCOE, Yadrav.

2. Assembly Language programming Basics.
Assembly language Programming
Assembly language is a low-level programming language for a
computer or other programmable device specific to a particular
computer architecture in contrast to most high-level
programming languages, which are generally portable across
multiple systems. Assembly language is converted into
executable machine code by a utility program referred to as an
assembler like NASM, MASM, etc.

3. Assembly Language programming Basics.
What is Assembly Language?
Each personal computer has a microprocessor that manages the computer's
arithmetical, logical, and control activities.
Each family of processors has its own set of instructions for handling various operations
such as getting input from keyboard, displaying information on screen and performing
various other jobs. These set of instructions are called 'machine language instructions'.
A processor understands only machine language instructions, which are strings of 1's
and 0's. However, machine language is too obscure and complex for using in software
development. So, the low-level assembly language is designed for a specific family of
processors that represents various instructions in symbolic code and a more
understandable form.

4. Assembly Language programming Basics.
Basic Features of PC Hardware
The main internal hardware of a PC consists of processor, memory, and registers. Registers are
processor components that hold data and address. To execute a program, the system copies it from
the external device into the internal memory. The processor executes the program instructions.
The fundamental unit of computer storage is a bit; it could be ON (1) or OFF (0) and a group of 8
related bits makes a byte on most of the modern computers.
The processor supports the following data sizes −
Word: a 2-byte data item
Double word: a 4-byte (32 bit) data item
Quad word: an 8-byte (64 bit) data item
Paragraph: a 16-byte (128 bit) area
Kilobyte: 1024 bytes
Megabyte: 1,048,576 bytes

5. Assembly Language programming Basics.
Binary Number System
Every number system uses positional notation, i.e., each position in which a digit is
written has a different positional value. Each position is power of the base, which is 2 for
binary number system, and these powers begin at 0 and increase by 1.
The following table shows the positional values for an 8-bit binary number, where all bits
are set ON.
The value of a binary number is based on the presence of 1 bits and their positional
value. So, the value of a given binary number is −
1 + 2 + 4 + 8 +16 + 32 + 64 + 128 = 255
which is same as 28 - 1.
Bit value 1 1 1 1 1 1 1 1
Position value as a power of
base 2
128 64 32 16 8 4 2 1
Bit number 7 6 5 4 3 2 1 0

6. Assembly Language programming Basics.
Hexadecimal Number System
Hexadecimal number system uses base 16. The digits in this system range from
0 to 15. By convention, the letters A through F is used to represent the
hexadecimal digits corresponding to decimal values 10 through 15.
Hexadecimal numbers in computing is used for abbreviating lengthy binary
representations. Basically, hexadecimal number system represents a binary data
by dividing each byte in half and expressing the value of each half-byte.

7. Assembly Language programming Basics.
Decimal number Binary representation Hexadecimal representation
0 0 0
1 1 1
2 10 2
3 11 3
4 100 4
5 101 5
6 110 6
7 111 7
8 1000 8
9 1001 9
10 1010 A
11 1011 B
12 1100 C
13 1101 D
14 1110 E
15 1111 F

8. Assembly Language programming Basics.
To convert a binary number to its hexadecimal equivalent, break it into
groups of 4 consecutive groups each, starting from the right, and write those
groups over the corresponding digits of the hexadecimal number.
Example − Binary number 1000 1100 1101 0001 is equivalent to
hexadecimal - 8CD1
To convert a hexadecimal number to binary, just write each hexadecimal digit
into its 4-digit binary equivalent.
Example − Hexadecimal number FAD8 is equivalent to binary - 1111 1010
1101 1000

9. Assembly Language programming Basics.
Binary Arithmetic
The following table illustrates four simple rules for binary addition −
(i) (ii) (iii) (iv)
1
0 1 1 1
+0 +0 +1 +1
=0 =1 =10 =11
Rules (iii) and (iv) show
a carry of a 1-bit into
the next left position.
Decimal Binary
60 00111100
+42 00101010
102 01100110
A negative binary value is expressed
in two's complement notation. According
to this rule, to convert a binary number to
its negative value is to reverse its bit
values and add 1.
Number 53 00110101
Reverse the bits 11001010
Add 1 00000001
Number -53 11001011
To subtract one value from another, convert the number being
subtracted to two's complement format and add the numbers.

10. Assembly Language programming Basics.
Binary Arithmetic
The following table illustrates four simple rules for binary addition −
(i) (ii) (iii) (iv)
1
0 1 1 1
+0 +0 +1 +1
=0 =1 =10 =11
Rules (iii) and (iv) show
a carry of a 1-bit into
the next left position.
Decimal Binary
60 00111100
+42 00101010
102 01100110
A negative binary value is expressed
in two's complement notation. According
to this rule, to convert a binary number to
its negative value is to reverse its bit
values and add 1.
Number 53 00110101
Reverse the bits 11001010
Add 1 00000001
Number -53 11001011
To subtract one value from another, convert the number being
subtracted to two's complement format and add the numbers.

11. 8085 Microprocessor Addressing Modes.
Addressing Modes in 8085
These are the instructions used to transfer the data from one register to another
register, from the memory to the register, and from the register to the memory without
any alteration in the content. Addressing modes in 8085 is classified into 5 groups −
Immediate addressing mode
In this mode, the 8/16-bit data is specified in the instruction itself as one of its operand.
For example: MVI K, 20F: means 20F is copied into register K.
Register addressing mode
In this mode, the data is copied from one register to another. For example: means
MOV K, B: means data in register B is copied to register K.

12. 8085 Microprocessor Addressing Modes.
Addressing Modes in 8085
Direct addressing mode
In this mode, the data is directly copied from the given address to the
register.
For example: LDB 5000K: means the data at address 5000K is copied
to register B.
Indirect addressing mode
In this mode, the data is transferred from one register to another by
using the address pointed by the register. For example: address MOV
K, B: means data is transferred from the memory address pointed by
the register to the register K.
Implied addressing mode
This mode doesn’t require any operand; the data is specified by the
opcode itself. For example: CMP.

13. 8085 Microprocessor Instruction Sets.
Control Instructions
Opcode Operand Meaning Explanation
NOP None No operation No operation is performed, i.e., the instruction is
fetched and decoded.
HLT None Halt and enter wait state The CPU finishes executing the current instruction
and stops further execution. An interrupt or reset is
necessary to exit from the halt state.
DI None Disable interrupts The interrupt enable flip-flop is reset and all the
interrupts are disabled except TRAP.
EI None Enable interrupts The interrupt enable flip-flop is set and all the
interrupts are enabled.
RIM None Read interrupt mask This instruction is used to read the status of
interrupts 7.5, 6.5, 5.5 and read serial data input bit.
SIM None Set interrupt mask This instruction is used to implement the interrupts
7.5, 6.5, 5.5, and serial data output.
Instruction sets are instruction codes to perform some task. It is classified into five categories

14. 8085 Microprocessor Instruction Sets.
Logical Instructions
Opcode Operand Meaning Explanation
CMP
R
M
Compare the register or
memory with the accumulator
The contents of the operand (register or
memory) are M compared with the contents of
the accumulator.
CPI 8-bit data
Compare immediate with the
accumulator
The second byte data is compared with the
contents of the accumulator.
ANA
R
M
Logical AND register or
memory with the accumulator
The contents of the accumulator are logically
AND with M the contents of the register or
memory, and the result is placed in the
accumulator.
ANI 8-bit data
Logical AND immediate with
the accumulator
The contents of the accumulator are logically
AND with the 8-bit data and the result is placed
in the accumulator.
XRA
R
M
Exclusive OR register or
memory with the accumulator
The contents of the accumulator are Exclusive
OR with M the contents of the register or
memory, and the result is placed in the
accumulator.

15. 8085 Microprocessor Instruction Sets.
Logical Instructions
Opcode Operand Meaning Explanation
XRI 8-bit data
Exclusive OR immediate with
the accumulator
The contents of the accumulator are Exclusive
OR with the 8-bit data and the result is placed in
the accumulator.
ORA
R
M
Logical OR register or
memory with the accumulator
The contents of the accumulator are logically OR
with M the contents of the register or memory,
and result is placed in the accumulator.
ORI 8-bit data
Logical OR immediate with
the accumulator
The contents of the accumulator are logically OR
with the 8-bit data and the result is placed in the
accumulator.
RLC None Rotate the accumulator left
Each binary bit of the accumulator is rotated left
by one position. Bit D7 is placed in the position
of D0 as well as in the Carry flag. CY is modified
according to bit D7.
RRC None Rotate the accumulator right
Each binary bit of the accumulator is rotated
right by one position. Bit D0 is placed in the
position of D7 as well as in the Carry flag. CY is
modified according to bit D0.

16. 8085 Microprocessor Instruction Sets.
Logical Instructions
Opcode Operand Meaning Explanation
RAL None
Rotate the accumulator left
through carry
Each binary bit of the accumulator is rotated left
by one position through the Carry flag. Bit D7 is
placed in the Carry flag, and the Carry flag is
placed in the least significant position D0. CY is
modified according to bit D7.
RAR None
Rotate the accumulator right
through carry
Each binary bit of the accumulator is rotated right
by one position through the Carry flag. Bit D0 is
placed in the Carry flag, and the Carry flag is
placed in the most significant position D7. CY is
modified according to bit D0.
CMA None
Complement accumulator The contents of the accumulator are
complemented. No flags are affected.
CMC None
Complement carry The Carry flag is complemented. No other flags
are affected.
STC None Set Carry Set Carry

17. 8085 Microprocessor Instruction Sets.
Branching Instructions
Opcode Operand Meaning Explanation
JMP
16-bit
address
Jump
uncondition
ally
The program sequence is
transferred to the memory
address given in the operand.
16-bit
address
Jump
conditionall
y
The program sequence is
transferred to the memory
address given in the operand
based on the specified flag of
the PSW.
Opcode Description Flag Status
JC Jump on Carry CY=1
JNC Jump on no Carry CY=0
JP Jump on positive S=0
JM Jump on minus S=1
JZ Jump on zero Z=1
JNZ Jump on no zero Z=0
JPE Jump on parity even P=1
JPO Jump on parity odd P=0

18. 8085 Microprocessor Instruction Sets.
Branching Instructions
Opcode Operand Meaning Explanation
16-bit
address
Unconditional
subroutine call
The program sequence is
transferred to the memory
address given in the operand.
Before transferring, the address
of the next instruction after
CALL is pushed onto the stack.
RET None Return from
subroutine
unconditionally
The program sequence is
transferred from the subroutine
to the calling program.
Opcode Description Flag Status
CC Call on Carry CY=1
CNC Call on no Carry CY=0
CP Call on positive S=0
CM Call on minus S=1
CZ Call on zero Z=1
CNZ Call on no zero Z=0
CPE Call on parity even P=1
CPO Call on parity odd P=0

19. 8085 Microprocessor Instruction Sets.
Branching Instructions
Opcode Operand Meaning Explanation
None
Return from
subroutine
conditionally
The program sequence is transferred
from the subroutine to the calling
program based on the specified flag of
the PSW and the program execution
begins at the new address.
PCHL None Load the
program
counter with
HL contents
The contents of registers H & L are copied
into the program counter. The contents of
H are placed as the high-order byte and
the contents of L as the low order byte.
Opcode Description Flag
Status
RC Return on Carry CY=1
RNC Return on no Carry CY=0
RP Return on positive S=0
RM Return on minus S=1
RZ Return on zero Z=1
RNZ Return on no zero Z=0
RPE Return on parity even P=1
RPO Return on parity odd P=0

20. 8085 Microprocessor Instruction Sets.
Branching Instructions
Opcode Operand Meaning Explanation
RST 0-7 Restart The RST instruction is used as software instructions in a
program to transfer the program execution to one of the
following eight locations.
Instruction Restart Address
RST 0 0000H
RST 1 0008H
RST 2 0010H
RST 3 0018H
RST 4 0020H
RST 5 0028H
RST 6 0030H
RST 7 0038H

21. 8085 Microprocessor Instruction Sets.
Branching Instructions
Opcode Operand Meaning Explanation
RST 0-7 Restart The 8085 has additionally 4 interrupts, which can generate
RST instructions internally and doesn’t require any external
hardware. Following are those instructions and their
Restart addresses −
Interrupt Restart Address
TRAP 0024H
RST 5.5 002CH
RST 6.5 0034H
RST 7.5 003CH

22. 8085 Microprocessor Instruction Sets.
Arithmetic Instructions
Opcode Operand Meaning Explanation
ADD
R
M
Add register or memory, to
the accumulator
The contents of the register or memory are added to the
contents of the accumulator and the result is stored in the
accumulator.
Example − ADD K.
ADC
R
M
Add register to the
accumulator with carry
The contents of the register or memory & M the Carry flag
are added to the contents of the accumulator and the result is
stored in the accumulator.
Example − ADC K
ADI 8-bit data
Add the immediate to the
accumulator
The 8-bit data is added to the contents of the accumulator
and the result is stored in the accumulator.
Example − ADI 55K
ACI 8-bit data
Add the immediate to the
accumulator with carry
The 8-bit data and the Carry flag are added to the contents of
the accumulator and the result is stored in the accumulator.
Example − ACI 55K
LXI
Reg. pair,
16bit data
Load the register pair
immediate
The instruction stores 16-bit data into the register pair
designated in the operand.
Example − LXI K, 3025M
DAD Reg. pair
Add the register pair to H
and L registers
The 16-bit data of the specified register pair are added to the
contents of the HL register.
Example − DAD K

23. 8085 Microprocessor Instruction Sets.
Arithmetic Instructions
Opcode Operand Meaning Explanation
SUB
R
M
Subtract the register or the
memory from the
accumulator
The contents of the register or the memory are subtracted
from the contents of the accumulator, and the result is stored
in the accumulator.
Example − SUB K
SBB
R
M
Subtract the source and
borrow from the
accumulator
The contents of the register or the memory & M the Borrow
flag are subtracted from the contents of the accumulator and
the result is placed in the accumulator.
Example − SBB K
SUI 8-bit data
Subtract the immediate from
the accumulator
The 8-bit data is subtracted from the contents of the
accumulator & the result is stored in the accumulator.
Example − SUI 55K
XCHG None
Exchange H and L with D
and E
The contents of register H are exchanged with the contents
of register D, and the contents of register L are exchanged
with the contents of register E.
Example − XCHG
INR
R
M
Increment the register or the
memory by 1
The contents of the designated register or the memory are
incremented by 1 and their result is stored at the same place.
Example − INR K

24. 8085 Microprocessor Instruction Sets.
Arithmetic Instructions
Opcode Operand Meaning Explanation
INX R Increment register pair by 1
The contents of the designated register pair are incremented
by 1 and their result is stored at the same place.
Example − INX K
DCR
R
M
Decrement the register or
the memory by 1
The contents of the designated register or memory are
decremented by 1 and their result is stored at the same
place.
Example − DCR K
DCX R
Decrement the register pair
by 1
The contents of the designated register pair are decremented
by 1 and their result is stored at the same place.
Example − DCX K
DAA None Decimal adjust accumulator
The contents of the accumulator are changed from a binary
value to two 4-bit BCD digits.
If the value of the low-order 4-bits in the accumulator is
greater than 9 or if AC flag is set, the instruction adds 6 to the
low-order four bits.
If the value of the high-order 4-bits in the accumulator is
greater than 9 or if the Carry flag is set, the instruction adds 6
to the high-order four bits.
Example − DAA

25. 8085 Microprocessor Instruction Sets.
Data Transfer Instructions
Opcode Operand Meaning Explanation
MOV
Rd, Sc
M, Sc
Dt, M
Copy from the source (Sc) to the
destination(Dt)
This instruction copies the contents of the source
register into the destination register without any
alteration.
Example − MOV K, L
MVI
Rd, data
M, data
Move immediate 8-bit
The 8-bit data is stored in the destination register or
memory.
Example − MVI K, 55L
LDA
16-bit
address
Load the accumulator
The contents of a memory location, specified by a 16-
bit address in the operand, are copied to the
accumulator.
Example − LDA 2034K
LDAX
B/D Reg.
pair
Load the accumulator indirect
The contents of the designated register pair point to a
memory location. This instruction copies the contents
of that memory location into the accumulator.
Example − LDAX K
LXI
Reg. pair,
16-bit data
Load the register pair immediate
The instruction loads 16-bit data in the register pair
designated in the register or the memory.
Example − LXI K, 3225L

26. 8085 Microprocessor Instruction Sets.
Data Transfer Instructions
Opcode Operand Meaning Explanation
LHLD
16-bit
address
Load H and L registers direct
The instruction copies the contents of the memory
location pointed out by the address into register L and
copies the contents of the next memory location into
register H. Example − LHLD 3225K
STA
16-bit
address
16-bit address
The contents of the accumulator are copied into the
memory location specified by the operand.
This is a 3-byte instruction, the second byte specifies
the low-order address and the third byte specifies the
high-order address. Example − STA 325K
STAX
16-bit
address
Store the accumulator indirect
The contents of the accumulator are copied into the
memory location specified by the contents of the
operand. Example − STAX K
SHLD
16-bit
address Store H and L registers direct
The contents of register L are stored in the memory
location specified by the 16-bit address in the operand
and the contents of H register are stored into the next
memory location by incrementing the operand.
This is a 3-byte instruction, the second byte specifies
the low-order address and the third byte specifies the
high-order address. Example − SHLD 3225K

27. 8085 Microprocessor Instruction Sets.
Data Transfer Instructions
Opcode Operand Meaning Explanation
XCHG None Exchange H and L with D and E
The contents of register H are exchanged with the
contents of register D, and the contents of register L
are exchanged with the contents of register E.
Example − XCHG
SPHL None
Copy H and L registers to the stack
pointer
The instruction loads the contents of the H and L
registers into the stack pointer register. The contents of
the H register provide the high-order address and the
contents of the L register provide the low-order
address.
Example − SPHL
XTHL None Exchange H and L with top of stack
The contents of the L register are exchanged with the
stack location pointed out by the contents of the stack
pointer register.
The contents of the H register are exchanged with the
next stack location (SP+1).
Example − XTHL

28. 8085 Microprocessor Instruction Sets.
Data Transfer Instructions
Opcode Operand Meaning Explanation
PUSH Reg. pair
Push the register pair onto the
stack
The contents of the register pair designated in the operand are
copied onto the stack in the following sequence.
The stack pointer register is decremented and the contents of the
high order register (B, D, H, A) are copied into that location.
The stack pointer register is decremented again and the contents
of the low-order register (C, E, L, flags) are copied to that
location. Example − PUSH K
POP Reg. pair
Pop off stack to the register
pair
The contents of the memory location pointed out by the stack
pointer register are copied to the low-order register (C, E, L,
status flags) of the operand.
The stack pointer is incremented by 1 and the contents of that
memory location are copied to the high-order register (B, D, H, A)
of the operand. The stack pointer register is again incremented
by 1. Example − POPK
OUT
8-bit port
address
Output the data from the
accumulator to a port with 8bit
address
The contents of the accumulator are copied into the I/O port
specified by the operand.
Example − OUT K9L
IN
8-bit port
address
Input data to accumulator from
a port with 8-bit address
The contents of the input port designated in the operand are read
and loaded into the accumulator. Example − IN5KL