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MOV Instruction MOV destination, source ; copy source to dest. MOV A,#55H ;load value 55H into reg. A MOV R0,A ;copy contents of A into R0 ;(now A=R0=55H)

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Presentation on theme: "MOV Instruction MOV destination, source ; copy source to dest. MOV A,#55H ;load value 55H into reg. A MOV R0,A ;copy contents of A into R0 ;(now A=R0=55H)"— Presentation transcript:

1 MOV Instruction MOV destination, source ; copy source to dest. MOV A,#55H ;load value 55H into reg. A MOV R0,A ;copy contents of A into R0 ;(now A=R0=55H) MOV R1,A ;copy contents of A into R1 ;(now A=R0=R1=55H) MOV R2,A ;copy contents of A into R2 ;(now A=R0=R1=R2=55H) MOV R3,#95H ;load value 95H into R3 ;(now R3=95H) MOV A,R3 ;copy contents of R3 into A ;now A=R3=95H

2 Notes on Programming Value (proceeded with #) can be loaded directly to registers A, B, or R0 – R7 –MOV R5, #0F9H If values 0 to F moved into an 8-bit register, the rest assumed all zeros –MOV A, #5 A too large value causes an error –MOV A, #7F2H

3 ADD Instruction ADD A, source ;ADD the source ; operand ;to the accumulator MOV A, #25H ;load 25H into A MOV R2,#34H ;load 34H into R2 ADD A,R2 ;add R2 to accumulator ;(A = A + R2)

4 ADD Instruction and PSW

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7 Structure of Assembly Language ORG 0H ;start (origin) at location 0 MOV R5,#25H ;load 25H into R5 MOV R7,#34H ;load 34H into R7 MOV A,#0 ;load 0 into A ADD A,R5 ;add contents of R5 to A ;now A = A + R5 ADD A,R7 ;add contents of R7 to A ;now A = A + R7 ADD A,#12H ;add to A value 12H ;now A = A + 12H HERE: SJMP HERE ;stay in this loop END ;end of asm source file

8 Data Types & Directives ORG 500H DATA1: DB 28 ;DECIMAL (1C in Hex) DATA2: DB B ;BINARY (35 in Hex) DATA3: DB 39H ;HEX ORG 510H DATA4: DB “2591” ; ASCII NUMBERS ORG 518H DATA6: DB “My name is Joe” ;ASCII CHARACTERS

9 Access RAM Locations Using Register Names

10 Access RAM Locations Using Addresses

11 Switch Register Banks

12 Pushing onto Stack

13 Popping from Stack

14 Stack & Bank 1 Conflict

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16 Arithmetic Instructions and Programs

17 Outlines Range of numbers in 8051 unsigned data Addition & subtraction instructions for unsigned data BCD system of data representation Packed and unpacked BCD data Addition & subtraction on BCD data Range of numbers in 8051 signed data Signed data arithmetic instructions Carry & overflow problems & corrections

18 Addition of Unsigned Numbers ADDA, source; A = A + source

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20 Addition of Individual Bytes

21 ADDC & Addition of 16-bit Numbers 1 3CE7 3B8D

22 BCD Number System Unpacked BCD: 1 byte Packed BCD: 4 bits

23 Adding BCD Numbers & DA Instruction MOVA,#17H ADDA,#28H MOVA,#47H;A=47H first BCD operand MOVB,#25H;B=25 second BCD operand ADDA,B;hex (binary) addition (A=6CH) DAA;adjust for BCD addition (A=72H) HEXBCD AC=

24 Example

25 Subtraction of Unsigned Numbers SUBBA, source; A = A – source – CY SUBB when CY = 0 –Take 2’s complement of subtraend (source) –Add it to minuend –Invert carry

26 Example (Positive Result)

27 Example (Negative Result)

28 SUBB When CY = 1 For multibyte numbers

29 Multiplication of Unsigned Numbers MULAB; A  B, place 16-bit result in B and A MOVA,#25H;load 25H to reg. A MOVB,#65H;load 65H in reg. B MULAB;25H * 65H = E99 where ;B = 0EH and A = 99H Table 6-1:Unsigned Multiplication Summary (MUL AB) MultiplicationOperand 1Operand 2Result byte  byteABA=low byte, B=high byte

30 Division of Unsigned Numbers DIVAB; divide A by B MOVA,#95H;load 95 into A MOVB,#10H;load 10 into B DIVAB;now A = 09 (quotient) and ;B = 05 (remainder) Table 6-2:Unsigned Division Summary (DIV AB) DivisionNumeratorDenominatorQuotientRemainder byte / byteABAB

31 Example ( 1 of 2 )

32 Example ( 2 of 2 )

33 Signed 8-bit Operands Convert to 2’s complement –Write magnitude of number in 8-bit binary (no sign) –Invert each bit –Add 1 to it

34 Example

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37 Byte-sized Signed Numbers Ranges DecimalBinaryHex ….………… FE FF …………… F

38 Overflow in Signed Number Operations

39 When Is the OV Flag Set? Either: there is a carry from D6 to D7 but no carry out of D7 (CY = 0) Or: there is a carry from D7 out (CY = 1) but no carry from D6 to D7

40 Example

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