Lecture 3Arithmetic, Logic Instructions,

Prof. S. M. Lutful KabirIICT, BUETSession: April, 2011

ICT 6641: Advanced Embedded SystemICT 6641: Advanced Embedded System

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Review on Lecture-2Review on Lecture-2

• Branching and Looping• Loop inside loop• Different types of JMP Instructions• Different types of CALL Instructions• Stack and Stack Pointer• The PUSH and POP Instruction• The Subroutine Call and Returning From it • Pipelining• Instruction Cycle Time for the AVR• Calculation of Delay Time

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Review on Lecture-2 (continued)Review on Lecture-2 (continued)

• AVR I/O Programming• I/O ports and their functions• The Role of DDRx Registers• Outputting Data to a Port• Reading Data from a Port• Built-in pull-up resistors with the input pins• Duel Role of Ports• I/O Bit Manipulation Programming• I/O Registers• SBI, CBI, SBIC and SBIS instructions

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Addition of Unsigned NumbersAddition of Unsigned Numbers

• In AVR, the add operation uses two general purpose registers as its input and the result is stored in first (left) Register

• The format is ADD Rd, Rr ; Rd=Rd+Rr

• ADD could change any of the Z, C, N, V, H and S bits of the status register depending of the operands involved

• Notice that none of the AVR Addition instruction direct memory access; to add memory location we should first load it to any of the registers R0-R31

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How the status bits are affected How the status bits are affected during additionduring addition

LDI R21, 0xF5 Solution :LDI R22, 0x0B F5H 1111 0101ADD R21, R22 0BH 0000 1011

100H 10000 0000• After the addition, register R21 contains 00 and the flags are

as follows,• C = 1, because there is a carry out of D7• Z = 1, because the result in the destination register (R21) is zero• H = 1, because there is a Carry from D3 to D4

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Addition using Data from the Addition using Data from the Memory LocationMemory Location

• Assume that the memory location 0x400 has the value of 33H. Write a program to add the content of memory location 0x400 and the data 55H

Solution:LDS R2, 0x400 ; R2 = 33H (location of 0x400 of RAM)LDI R21, 0x55 ; R21 = 55HADD R21, R2 ; R21 = R21 + R2 = 55H + 33H = 88H,

C = 0, Z = 0, H = 0

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ADC and Addition of 16 bitsADC and Addition of 16 bits• When adding two 16 bit operands, we need to be concerned with the

propagation of carry from the lower byte to the higher byte • This is called multi-byte addition to distinguish it from addition of

individual bytes• The instruction ADC (ADD with Carry) is used on such occasions• For example, let us see the addition of 3CE7H+3B8DH

13C E73B 8D78 74

• Assume that R1 = 8D, R2 = 3B, R3 = E7 and R4 = 3C,ADD R3, R1 ; R3 = R3 + R1 = E7 + 8D = 74 and C = 1 ADC R4, R2 ; R4 = R4 + R2 + Carry = 3C + 3B + 1 = 78

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Subtraction of unsigned numbersSubtraction of unsigned numbers• In many microprocessors, there are two different instructions:

SUB and SUBB• In AVR we have five instructions for subtraction: SUB, SBC,

SUBI, SBCI and SBIW• Their formats are:

– SUB Rd, Rr ; Rd = Rd - Rr– SBC Rd, Rr ; Rd = Rd - Rr - C– SUBI Rd, K ; Rd = Rd - K– SBIC Rd, K ; Rd = Rd - K - C

– SBIW Rd:Rd+1, K ; Rd + 1 : Rd = Rd + 1 : Rd - K S• In AVR, SBC and SBCI are subtraction with borrow. In AVR we

use Carry flag for borrow and so SBC (SuB with Carry)

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SUB Rd, Rr [Rd = Rd – Rr]SUB Rd, Rr [Rd = Rd – Rr]

• In subtraction, AVR microcontroller (in fact all modern CPUs) use the 2’s complement method

• One can summarize the steps of the hardware of the CPU in executing the SUB instruction for unsigned numbers as follows:

• 1. Take the 2’s complement of subtrahend (left side operand)• 2. Add with the minuend (right side operator)• 3. Invert the carry

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Example of SUB instructionExample of SUB instruction• Show the steps involved in the following:• LDI R20, 0x23• LDI R21, 0x3F• SUB R21, R20Solution

R21 = 3F 0011 1111 0011 1111R20 = 23 0010 0011 1101 1101 (2’s complement) 1C 1 0001 1100

• C = 0 and D7=N=0• The flags will be set as follows: C=0 and N=0 (Notice that there

is a carry but C flag will be set to 0, [this will be discussed later]• The programmer will look into N (or C) to determine +ve or -ve

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If the result is negativeIf the result is negative

• It is already mentioned that if N=0 (or C = 0) the result is positive

• But if N = 1 (or C = 1), the result will be negative and the destination will contain the 2’s complement of the result. NEG (negate) command can be used to change it

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SUBI and SBIWSUBI and SBIW

• The other subtraction instructions are SUBI and SBIW, which subtracts an immediate (constant) value from a register

• The SBIW subtracts an immediate value in the range of 0-63 from a register pair and stores the result in the register pair

• Notice that only last 8 registers can be used for SBIW

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Subtraction of 18H from 2917H Subtraction of 18H from 2917H

LDI R25, 0x29LDI R24, 0x17SBIW R25:24, 0x18

• Notice that you should use SBIW Rd+1:Rd, K format• If you type SWIB Rd:Rd+1, K the assembler will assemble your

code as if you typed SBIW Rd+1:Rd, K

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SBC (Rd SBC (Rd Rd – Rr – C) Rd – Rr – C) Subtract with borrow (denoted by C)Subtract with borrow (denoted by C)

• This instruction is used for multi-byte numbers and will take care of borrow of the lower byte

• If the borrow flag is set to one (C=1) prior to executing the SBC instruction, this operation also subtracts 1 from the result

• Example: 1 LDI R26, 0x6227 62H LDI R27, 0x9612 96H LDI R28, 0x9614 CCH LDI R29, 0x12

SUB R26, R28 --> C=1, N=1

SBC R27, R29

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The C flag in subtraction of AVRThe C flag in subtraction of AVR

• Notice that AVR is like other CPUs such as the x86 and the 8051 when it comes to the carry flag in subtract operations. In the AVR, after subtract operations

• In AVR after the subtract operations, the carry is inverted by the CPU itself and we examine the C flag to see if the result is positive or negative

• That means that if C=0, the result is positive and if C=1, the result is negative

• Notice that the CPU does not invert the carry flag after the ADD instruction

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Multiplication of unsigned numbersMultiplication of unsigned numbers

• The AVR has several instructions dedicated to multiplication• First we shall discuss MUL instruction and other are similar to

MUL but are used for signed numbers• MUL is byte by byte multiply instruction• In byte by byte multiply instruction, operands must be in

registers.• After multiplication the unsigned product is placed in R1 (high

byte) and R0 (low byte)• Notice that if any of the operands is selected from R0 or R1,

the result will overwrite those registers after multiplication

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Multiplication SummaryMultiplication Summary

-------------------------------------------------------------------------------------------------------Multiplication Application Byte1 Byte2 High byte Low byte

of result of result

-------------------------------------------------------------------------------------------------------

MUL Rd, Rr Unsigned numbers Rd Rr R1 R0

MULS Rd, Rr Signed numbers Rd Rr R1 R0

MULSU Rd, Rr Unsigned number with Rd Rr R1 R0 signed number

-------------------------------------------------------------------------------------------------------

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Division of unsigned numbersDivision of unsigned numbers

• AVR has no instruction for divide operation• We can write a program by repeated subtraction• In dividing a byte by a byte the numerator is placed in a

register and the denominator is subtracted from it repeatedly.

• The quotient is the number of times and the remainder is the register upon completion

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An example of DivisionAn example of Division

.DEF NUM=R20

.DEF DENOMINATOR = R21

.DEF QUOTIENT = R22LDI NUM, 95LDI DENOMINATOR, 10CLR QUOTIENT

L1: INC QUOTIENTSUB NUM, DENOMINATORBRCC L1

DEC QUOTIENT ; extra increment is reducedADD NUM, DENOMINATOR ; NUM is now the remainder

HERE: JMP HERE

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AssignmentAssignment

• Analyze the example 5-8 of page 168

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Signed Number Concepts and Signed Number Concepts and Arithmetic OperationsArithmetic Operations

• All data items used so far have been unsigned numbers meaning that the entire 8-bit operand was used for the magnitude

• To represent a number as signed, the MSB is set aside for sign• The sign is represented by 0 for positive (+ve) numbers and 1

for negative (-ve) numbers• The steps in representing a negative number (in 2’s

Complement)– Represent the number in 8-bit binary– Invert the digits– Add 1 with the inverted number

• -127 is represented by 81H, -34H by CCH, -5 as FBH

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Overflow problem in signed number Overflow problem in signed number operations operations

• When using signed numbers, a serious problem sometimes arises that must be dealt with

• This is the Overflow problem, the AVR indicates the existence of an error by raising the V (overflow) flag, but it is upto the programmer to care of the erroneous result

• If the result of an operation on signed numbers is too large for the register, an overflow occurs, V flag is set

• Let us see the following operation +96 0110 0000

+ +70 0100 0110The limit is +127 +166 1010 0110 (-90) INVALID RESULT N=1, V=1

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When is the V flag set?When is the V flag set?

• In 8-bit signed operations, V is set to 1 if either of the following conditions occurs:– There is a carry from D6 to D7 but no carry out of D7 (C=0)– There is a carry from D7 out (C=1) but no carry from D6 to D7

• In other words, the overflow flag is set the overflow flag is set to 1 if there is a carry from D6 to D7 or from D7 out but not the both

• It means that if there are carry both from D6 to D7 and from D7 out, V=0

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Some example of OverflowSome example of Overflow• Carry from D6 to D7 only, no carry from D7 out

+96 0110 0000 + +70 0100 0110 +166 1010 0110 (N=0,C=0,V=1) result is -90 WRONG• Carry from D7 out only, no carry from D6 to D7

- 128 1000 0000+ - 2 1111 1110

- 130 1 0111 1110 ( N=0,C=1,V=1) result is +126 WRONG• Carry both from D6 to D7 and D7 out

- 126 1000 0010+ - 2 1111 1110

- 128 1 1000 0000 ( N=1, C=1, V=0)

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Further Consideration on V flagFurther Consideration on V flag• In the ADD instruction there are two different conditions,

either the operands have the same sign or the signs of the operands are different

• When we ADD operands of different signs, the absolute values of the operands are less than absolute values of one of the operands, so overflow can not happen

• Overflow can only happen when operands of same sign, it is possible that the result overflows the limit of the register

• In AVR, the equation of V flag is as follows:V = Rd7.Rr7.R7+ Rd7.Rr7.R7

• Where Rd7 and Rr7 are 7th bit of the operands and R7 is the 7th bit of the result

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Further Consideration on V flag Further Consideration on V flag (continued)(continued)

• We conclude that in any signed number addition, V flag indicates that whether the result is valid or not

• If V=1, the result is erroneous, if V=0, the result is valid• We can state emphatically that in unsigned number addition,

the V (overflow) flag must be monitored• In AVR, there are BRVC and BRVS instructions for the V flag

that allow us to correct the signed number error• We also have BRCC and BRCS instructions to check for C flag

and BRPL and BRMI for checking N flag

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Difference between N and S flagDifference between N and S flag

• As we mentioned before that the N flag represents D7 bit of the result

• If the result is positive, the N flag is zero and if the result is negative N flag is 1

• In operations of signed numbers, overflow is possible. Overflow corrupts the result and negates (make opposite) the D7 bit

• So if we ADD two positive numbers, in case of overflow, the N flag would be 1 showing that the result is negative! (whether actual or not)

• The S flag helps you to know the real sign of the actual result

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Logic and Compare InstructionLogic and Compare Instruction

• Apart from I/O and arithmetic instructions, logic instructions are some of the most widely used instructions

ANDAND Rd, Rr ; Rd = Rd AND RrANDI Rd, K ; Rd = Rd AND K

• The AND instructions can affect Z, S and N flags• The AND instructions are often used to mask certain bits of an

operand LDI R20, 0x35 35H = 0011 0101 ANDI R20, 0x0F 0FH = 0000 1111

0000 0101 [masking upper nibble]

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Logical OR instructionLogical OR instructionOR Rd, Rr ; Rd = Rd OR Rr

• This instruction will perform a logical OR operation on two operands and keep the result is left hand operator

• There are also ORI instruction to perform OR operation between the content of a register and an immediate value, the format is

ORI Rd, K ; Rd = Rd OR K• OR instructions affects the Z, S and N flags• The OR instructions can be used to set certain bits of an

operand to 1, for exampleLDI R20, 0x04ORI R20, 0x30 ; now R20 will be 34H

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An example of ORI and ANDIAn example of ORI and ANDI

• Assume that PB2 is used to control an outdoor light and PB5 to control a light inside a building. Write a program to turn “on” the outdoor light and turn “off” the inside one.

SBI DDRB, 2 ORI R20, 0b00000100SBI DDRB, 5 AND R20, 0b11011111IN R20, PORTB OUT PORTB, R20

• Notice that we read PORTB instead of PINB because we want to know the last value of PORTB, not the value of chip’s pin

• ORI makes bit2 to 1 and keeps other bits as it is• ANDI makes bit5 to 0 and keeps other bits as it is

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EX-OR instructionEX-OR instruction• EOR Rd, Rs ; Rd = Rd XOR Rs• This instruction performs EX-OR between two operands and

keeps the result is the left operand• The EX-OR will affect the Z, S and N flags• EX-OR can be used to check whether the values in two

registers are equal or notOVER : IN R20, PORTB

LDI R21, 0x45EOR R20, R21BRNE OVER

• Another widely used application of EX-OR is to toggle the bits of an operand; EOR R0, R20, and R20 is initialized with 0xFF

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Few Other Logical InstructionsFew Other Logical Instructions

• COM (Complement) – COM R20 ; 1’s complement• NEG (Negative) – NEG R20 ; 2’s complement• CP (Compare) – CP Rd, Rr ; Compare is same as

subtraction except that compare does not change the register• There is also compare with constant value – CP Rd, K

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Branch InstructionBranch Instruction

Instruction Meaning Condition

BREQ Branch if equal Z=1

BRNE Branch if not equal Z=0

BRSH Branch if same or higher C=0

BRLO Branch if lower C=1

BRLT Branch if less than (signed) S=1

BRGE Branch if greater than or equal (signed) S=0

BRVS Branch if overflow flag set V=1

BRVC Branch if overflow flag clear V=0

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Rotate and Shift InstructionsRotate and Shift Instructions

• ROR Rd ; Rotate Rd right through carry

• ROL Rd ; Rotate Rd left through carry

MSB LSB C

MSB LSBC

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Serializing DataSerializing Data

• Serializing data is a way of sending a byte of data one bit at a time through a single pin of microcontroller

• There are two ways to transfer a byte of data serially– Using one serial port. In using the serial port, programmers have very

limited control over the sequence of data transfer– The second method of data transfer is to transfer one bit at a time

and control the sequence of data and space between them in many new generation of devices such as LCD, ADC, ROM and the serial versions are becoming popular because they take less space on a printed circuit board

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Sending data serially under program Sending data serially under program controlcontrol

SBI DDRB, 1 JMP NEXTLDI R20, 0x41 ONE: SBI PORTB, 1CLC NEXT:LDI R16, 8 DEC R16SBI PORTB, 1 BRNE AGAIN

AGAIN: SBI PORTB, 1ROR R20BRCS ONE HERE: JMP HERECBI PORTB, 1

• Write a program to transfer the value 41H serially one bit at a time via pin PB1. Put one high at the start and end of the data send LSB first

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Bring Data SeriallyBring Data Serially

CBI DDRC, 1LDI R16, 8LDI R20, 0

AGAIN:SBIC PINC, 7SECSBIS PINC, 7CLCROR R20DEC R16BRNE AGAIN

HERE: JMP HERE

• Write a program to bring a byte of data serially through pin PC7 and save it in R20 register. The byte comes in LSB first

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Shift and Swap InstructionsShift and Swap Instructions

• There are three shift instructions in AVR– LSL Rd ; Logical shift left– LSR Rd ; Logical shift right– ASR Rd ; Arithmetic shift right keeping MSB constant

• Swap Instruction• Nibble swapping – left nibble goes to right and right to left

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End of Lecture 3