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Instruction set

The MSP430 instruction set consists of 27 core instructions. Additionally, it supports 24 emulated instructions. The core instructions have unique op-codes decoded by the CPU, while the emulated ones need assemblers and compilers to generate their mnemonics.

There are three core-instruction formats:

- Double operand;

- Single operand;

- Program flow control - Jump.

Byte, word and address instructions are accessed using the .B, .W or .A extensions. If the extension is omitted, the instruction is interpreted as a word instruction.

Double operand instructions

The double operand instruction is formatted as follows:

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
opcode S-Reg Ad B/W As D-Reg
Bit Description
15-12 opcode
11-8 S-Reg The working register used for the source operand (src)
7 Ad The addressing bits responsible for the addressing mode used for the destination operand (dst)
6 B/W Byte or word operation:B/W=0: word operation; B/W=1: byte operation
5-4 As The addressing bits responsible for the addressing mode used for the source operand (src)
3-0 D-Reg The working register used for the destination operand (dst)

The next table shows the double operand instructions that are not emulated.

Mnemonic Operation Description
Arithmetic instructions
ADD(.B or .W) src,dst src+dst→dst Add source to destination
ADDC(.B or .W) src,dst src+dst+C→dst Add source and carry to destination
DADD(.B or .W) src,dst src+dst+C→dst (dec) Decimal add source and carry to destination
SUB(.B or .W) src,dst dst+.not.src+1→dst Subtract source from destination
SUBC(.B or .W) src,dst dst+.not.src+C→dst Subtract source and not carry from destination
Logical and register control instructions
AND(.B or .W) src,dst src.and.dst→dst AND source with destination
BIC(.B or .W) src,dst .not.src.and.dst→dst Clear bits in destination
BIS(.B or .W) src,dst src.or.dst→dst Set bits in destination
BIT(.B or .W) src,dst src.and.dst Test bits in destination
XOR(.B or .W) src,dst src.xor.dst→dst XOR source with destination
Data instructions
CMP(.B or .W) src,dst dst-src Compare source to destination
MOV(.B or .W) src,dst src→dst Move source to destination

Depending on the double operand instruction result, the status bits may be affected. The following gives the conditions for setting and resetting the status bits.

Status bits
Mnemonic V N Z C
Arithmetic instructions
ADD(.B or .W) src,dst =1, Arithmetic overflow=0, otherwise =1, negative result=0, if positive =1, null result=0, otherwise =1, carry from result=0, if not
ADDC(.B or .W) src,dst =1, Arithmetic overflow=0, otherwise =1, negative result=0, if positive =1, null result=0, otherwise =1, carry from MSB result=0, if not
DADD(.B or .W) src,dst - =1, MSB=1=0, otherwise =1, null result=0, otherwise =1, result>99(99)
SUB(.B or .W) src,dst =1, Arithmetic overflow=0, otherwise =1, negative result=0, if positive =1, null result=0, otherwise =1, if no borrow=0, otherwise
SUBC(.B or .W) src,dst =1, Arithmetic overflow=0, otherwise =1, negative result=0, if positive =1, null result=0, otherwise =1, if no borrow=0, otherwise
Logical and register control instructions
AND(.B or .W) src,dst =0 =1, MSB result set=0, if not set =1, null result=0, otherwise =1, not zero=0, otherwise
BIC(.B or .W) src,dst - - - -
BIS(.B or .W) src,dst - - - -
BIT(.B or .W) src,dst =0 =1, MSB result set=0, otherwise =1, null result=0, otherwise =1, not zero=0, otherwise
XOR(.B or .W) src,dst =1, both operands negative =1, MSB result set=0, otherwise =1, null result,=0, otherwise =1, not zero=0, otherwise
Data instructions
CMP(.B or .W) src,dst =1, Arithmetic overflow=0, otherwise =1, src>=dst=0, src<dst =1, src=dst=0, otherwise =1, carry from MSB result=0, if not
MOV(.B or .W) src,dst - - - -

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
Syamthanda Reply
hey , can you please explain oxidation reaction & redox ?
Boitumelo Reply
hey , can you please explain oxidation reaction and redox ?
Boitumelo
for grade 12 or grade 11?
Sibulele
the value of V1 and V2
Tumelo Reply
advantages of electrons in a circuit
Rethabile Reply
we're do you find electromagnetism past papers
Ntombifuthi
what a normal force
Tholulwazi Reply
it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface
Sihle
what is physics?
Petrus Reply
what is the half reaction of Potassium and chlorine
Anna Reply
how to calculate coefficient of static friction
Lisa Reply
how to calculate static friction
Lisa
How to calculate a current
Tumelo
how to calculate the magnitude of horizontal component of the applied force
Mogano
How to calculate force
Monambi
a structure of a thermocouple used to measure inner temperature
Anna Reply
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
Amahle Reply
How is energy being used in bonding?
Raymond Reply
what is acceleration
Syamthanda Reply
a rate of change in velocity of an object whith respect to time
Khuthadzo
how can we find the moment of torque of a circular object
Kidist
Acceleration is a rate of change in velocity.
Justice
t =r×f
Khuthadzo
how to calculate tension by substitution
Precious Reply
hi
Shongi
hi
Leago
use fnet method. how many obects are being calculated ?
Khuthadzo
khuthadzo hii
Hulisani
how to calculate acceleration and tension force
Lungile Reply
you use Fnet equals ma , newtoms second law formula
Masego
please help me with vectors in two dimensions
Mulaudzi Reply
how to calculate normal force
Mulaudzi
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Source:  OpenStax, Teaching and classroom laboratories based on the “ez430” and "experimenter's board" msp430 microcontroller platforms and code composer essentials. OpenStax CNX. May 19, 2009 Download for free at http://cnx.org/content/col10706/1.3
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