• Data Transfer Instructions

Data Transfer, Addressing and Arithmetic

Computer Organization and Assembly Languages Yung-Yu Chuang

2006/11/6

with slides by Kip Irvine

Chapter overview

• Data Transfer Instructions

• Addition and Subtraction

• Data-Related Operators and Directives

• Indirect Addressing

• JMP and LOOP Instructions

Data transfer instructions

• Operand Types

• Instruction Operand Notation

• Direct Memory Operands

• MOV Instruction

• Zero & Sign Extension

• XCHG Instruction

• Direct-Offset Instructions

Operand types

• Three basic types of operands:

– Immediate – a constant integer (8, 16, or 32 bits)

• value is encoded within the instruction – Register – the name of a register

• register name is converted to a number and encoded within the instruction

– Memory – reference to a location in memory

• memory address is encoded within the

instruction, or a register holds the address of a memory location

Instruction operand notation Direct memory operands

• A direct memory operand is a named reference to storage in memory

• The named reference (label) is automatically dereferenced by the assembler

.data

var1 BYTE 10h, .code

mov al,var1 ; AL = 10h mov al,[var1] ; AL = 10h

alternate format

MOV instruction

• Move from source to destination. Syntax:

MOVdestination, source

• Source and destination have the same size

• No more than one memory operand permitted

• CS, EIP, and IP cannot be the destination

• No immediate to segment moves

MOV instruction

.data

count BYTE 100 wVal WORD 2 .code

mov bl,count mov ax,wVal mov count,al

mov al,wVal ; error mov ax,count ; error mov eax,count ; error

Your turn . . .

Explain why each of the following MOV statements are invalid:

.data

bVal BYTE 100 bVal2 BYTE ? wVal WORD 2 dVal DWORD 5 .code

mov ds,45 ; a.

mov esi,wVal ; b.

mov eip,dVal ; c.

mov 25,bVal ; d.

mov bVal2,bVal ; e.

Memory to memory

.data

var1 WORD ? var2 WORD ? .code

mov ax, var1 mov var2, ax

Copy smaller to larger

.data

count WORD 1 .code

mov ecx, 0 mov cx, count .data

signedVal SWORD -16 ; FFF0h .code

mov ecx, 0 ; mov ecx, 0FFFFFFFFh mov cx, signedVal

MOVZX and MOVSX instructions take care of extension for both sign and unsigned integers.

Zero extension

mov bl,10001111b

movzx ax,bl ; zero-extension When you copy a smaller value into a larger destination, the MOVZX instruction fills (extends) the upper half of the destination with zeros.

1 0 0 0 1 1 1 1

1 0 0 0 1 1 1 1

Source

Destination 0 0 0 0 0 0 0 0

0

The destination must be a register.

movzx r32,r/m8 movzx r32,r/m16 movzx r16,r/m8

Sign extension

mov bl,10001111b

movsx ax,bl ; sign extension

The MOVSX instruction fills the upper half of the destination with a copy of the source operand's sign bit.

1 0 0 0 1 1 1 1

1 0 0 0 1 1 1 1

Source

Destination 1 1 1 1 1 1 1 1

The destination must be a register.

MOVZX MOVSX

From a smaller location to a larger one

mov bx, 0A69Bh

movzx eax, bx ; EAX=0000A69Bh movzx edx, bl ; EDX=0000009Bh movzx cx, bl ; EAX=009Bh mov bx, 0A69Bh

movsx eax, bx ; EAX=FFFFA69Bh movsx edx, bl ; EDX=FFFFFF9Bh movsx cx, bl ; EAX=FF9Bh

LAHF SAHF

.data

saveflags BYTE ? .code

lahf

mov saveflags, ah ...

mov ah, saveflags sahf

XCHG Instruction

XCHGexchanges the values of two operands. At least one operand must be a register. No immediate operands are permitted.

.data

var1 WORD 1000h var2 WORD 2000h .code

xchg ax,bx ; exchange 16-bit regs xchg ah,al ; exchange 8-bit regs xchg var1,bx ; exchange mem, reg xchg eax,ebx ; exchange 32-bit regs xchg var1,var2 ; error 2 memory operands

Exchange two memory locations

.data

var1 WORD 1000h var2 WORD 2000h .code

mov ax, val1 xchg ax, val2 mov val1, ax

Direct-offset operands

.data

arrayB BYTE 10h,20h,30h,40h .code

mov al,arrayB+1 ; AL = 20h

mov al,[arrayB+1] ; alternative notation mov al,arrayB+3 ; AL = 40h

A constant offset is added to a data label to produce an effective address (EA). The address is dereferenced to get the value inside its memory location. (no range checking)

Direct-offset operands

.data

arrayW WORD 1000h,2000h,3000h arrayD DWORD 1,2,3,4

.code

mov ax,[arrayW+2] ; AX = 2000h mov ax,[arrayW+4] ; AX = 3000h

mov eax,[arrayD+4] ; EAX = 00000002h A constant offset is added to a data label to produce an effective address (EA). The address is dereferenced to get the value inside its memory location.

; will the following assemble and run?

mov ax,[arrayW-2] ; ??

mov eax,[arrayD+16] ; ??

Your turn. . .

Write a program that rearranges the values of three doubleword values in the following array as: 3, 1, 2.

.data

arrayD DWORD 1,2,3

• Step 2: Exchange EAX with the third array value and copy the value in EAX to the first array position.

•Step1: copy the first value into EAX and exchange it with the value in the second position.

mov eax,arrayD

xchg eax,[arrayD+4]

xchg eax,[arrayD+8]

mov arrayD,eax

Evaluate this . . .

• We want to write a program that adds the following three bytes:

.data

myBytes BYTE 80h,66h,0A5h

• What is your evaluation of the following code?

mov ax,myBytes add ax,[myBytes+1]

add ax,[myBytes+2]

• What is your evaluation of the following code?

mov al,myBytes add al,[myBytes+1]

add al,[myBytes+2]

Evaluate this . . .

.data

myBytes BYTE 80h,66h,0A5h

Yes: Move zero to BX before the MOVZX instruction.

• How about the following code. Is anything missing?

movzx ax,myBytes mov bl,[myBytes+1]

add ax,bx

mov bl,[myBytes+2]

add ax,bx ; AX = sum

Addition and Subtraction

• INC and DEC Instructions

• ADD and SUB Instructions

• NEG Instruction

• Implementing Arithmetic Expressions

• Flags Affected by Arithmetic

– Zero – Sign – Carry – Overflow

INC and DEC Instructions

• Add 1, subtract 1 from destination operand

– operand may be register or memory

• INC destination

• Logic: destination ← destination + 1

• DEC destination

• Logic: destination ← destination – 1

INC and DEC Examples

.data

myWord WORD 1000h

myDword DWORD 10000000h .code

inc myWord ; 1001h dec myWord ; 1000h inc myDword ; 10000001h mov ax,00FFh

inc ax ; AX = 0100h mov ax,00FFh

inc al ; AX = 0000h

Your turn...

Show the value of the destination operand after each of the following instructions executes:

.data

myByte BYTE 0FFh, 0 .code

mov al,myByte ; AL = mov ah,[myByte+1] ; AH =

dec ah ; AH =

inc al ; AL =

dec ax ; AX =

FFh 00h FFh 00h FEFF

ADD and SUB Instructions

•ADD destination, source

• Logic: destination ← destination + source

•SUB destination, source

• Logic: destination ← destination – source

• Same operand rules as for the MOV instruction

ADD and SUB Examples

.data

var1 DWORD 10000h var2 DWORD 20000h

.code ; ---EAX---

mov eax,var1 ; 00010000h add eax,var2 ; 00030000h add ax,0FFFFh ; 0003FFFFh add eax,1 ; 00040000h sub ax,1 ; 0004FFFFh

NEG (negate) Instruction

.data

valB BYTE -1 valW WORD +32767 .code

mov al,valB ; AL = -1

neg al ; AL = +1

neg valW ; valW = -32767 Reverses the sign of an operand. Operand can be a register or memory operand.

Suppose AX contains –32,768 and we apply NEG to it.

Will the result be valid?

Implementing Arithmetic Expressions

Rval DWORD ? Xval DWORD 26 Yval DWORD 30 Zval DWORD 40 .code

mov eax,Xval

neg eax ; EAX = -26 mov ebx,Yval

sub ebx,Zval ; EBX = -10 add eax,ebx

mov Rval,eax ; -36

HLL compilers translate mathematical expressions into assembly language. You can do it also. For example:

Rval = -Xval + (Yval – Zval)

Your turn...

mov ebx,Yval neg ebx

add ebx,Zval mov eax,Xval sub ebx

mov Rval,eax

Translate the following expression into assembly language.

Do not permit Xval, Yval, or Zval to be modified:

Rval = Xval - (-Yval + Zval) Assume that all values are signed doublewords.

Flags Affected by Arithmetic

• The ALU has a number of status flags that

reflect the outcome of arithmetic (and bitwise) operations

– based on the contents of the destination operand

• Essential flags:

– Zero flag – destination equals zero – Sign flag – destination is negative – Carry flag – unsigned value out of range – Overflow flag – signed value out of range

• The MOV instruction never affects the flags.

Concept Map

status flags

ALU conditional

jumps

branching logic arithmetic &

bitwise operations

part of

used by provide attached to

affect

CPU

executes

executes

Zero Flag (ZF)

mov cx,1

sub cx,1 ; CX = 0, ZF = 1 mov ax,0FFFFh

inc ax ; AX = 0, ZF = 1 inc ax ; AX = 1, ZF = 0

Whenever the destination operand equals Zero, the Zero flag is set.

A flag is set when it equals 1.

A flag is clear when it equals 0.

Sign Flag (SF)

mov cx,0

sub cx,1 ; CX = -1, SF = 1 add cx,2 ; CX = 1, SF = 0 The Sign flag is set when the destination operand is negative. The flag is clear when the destination is positive.

The sign flag is a copy of the destination's highest bit:

mov al,0

sub al,1 ; AL=11111111b, SF=1 add al,2 ; AL=00000001b, SF=0

Carry Flag (CF)

The Carry flag is set when the result of an operation generates an unsigned value that is out of range (too big or too small for the destination operand).

mov al,0FFh

add al,1 ; CF = 1, AL = 00

; Try to go below zero:

mov al,0

sub al,1 ; CF = 1, AL = FF

In the second example, we tried to generate a negative value. Unsigned values cannot be negative, so the Carry flag signaled an error condition.

Carry Flag (CF)

• Addition and CF: copy carry out of MSB to CF

• Subtraction and CF: copy inverted carry out of MSB to CF

• INC/DEC do not affect CF

• Applying NEG to a nonzero operand sets CF

Your turn . . .

mov ax,00FFh

add ax,1 ; AX= SF= ZF= CF=

sub ax,1 ; AX= SF= ZF= CF=

add al,1 ; AL= SF= ZF= CF=

mov bh,6Ch

add bh,95h ; BH= SF= ZF= CF=

mov al,2

sub al,3 ; AL= SF= ZF= CF=

For each of the following marked entries, show the values of the destination operand and the Sign, Zero, and Carry flags:

0100h 0 0 0 00FFh 0 0 0 00h 0 1 1 01h 0 0 1

FFh 1 0 1

Overflow Flag (OF)

The Overflow flag is set when the signed result of an operation is invalid or out of range.

; Example 1 mov al,+127

add al,1 ; OF = 1, AL = ??

; Example 2

mov al,7Fh ; OF = 1, AL = 80h add al,1

The two examples are identical at the binary level because 7Fh equals +127. To determine the value of the destination operand, it is often easier to calculate in hexadecimal.

A Rule of Thumb

• When adding two integers, remember that the Overflow flag is only set when . . .

– Two positive operands are added and their sum is negative

– Two negative operands are added and their sum is positive

What will be the values of OF flag?

mov al,80h

add al,92h ; OF = mov al,-2

add al,+127 ; OF =

Your turn . . .

mov al,-128

neg al ; CF = OF =

mov ax,8000h

add ax,2 ; CF = OF = mov ax,0

sub ax,2 ; CF = OF = mov al,-5

sub al,+125 ; CF = OF =

What will be the values of the Carry and Overflow flags after each operation?

0 1

0 0

1 0

0 1

Signed/Unsigned Integers: Hardware Viewpoint

• All CPU instructions operate exactly the same on signed and unsigned integers

• The CPU cannot distinguish between signed and unsigned integers

• YOU, the programmer, are solely responsible for using the correct data type with each instruction

Overflow/Carry Flags: Hardware Viewpoint

• How the ADD instruction modifies OF and CF:

– CF = (carry out of the MSB)

– OF = (carry out of the MSB) XOR (carry into the MSB)

• How the SUB instruction modifies OF and CF:

– NEG the source and ADD it to the destination – CF = INVERT (carry out of the MSB)

– OF = (carry out of the MSB) XOR (carry into the MSB)

Auxiliary Carry (AC) flag

• AC indicates a carry or borrow of bit 3 in the destination operand.

• It is primarily used in binary coded decimal (BCD) arithmetic.

mov al, oFh

add al, 1 ; AC = 1

Parity (PF) flag

• PF is set when LSB of the destination has an even number of 1 bits.

mov al, 10001100b

add al, 00000010b ; AL=10001110, PF=1 sub al, 10000000b ; AL=00001110, PF=0

Data-Related Operators and Directives

• OFFSET Operator

• PTR Operator

• TYPE Operator

• LENGTHOF Operator

• SIZEOF Operator

• LABEL Directive

OFFSET Operator

• OFFSET returns the distance in bytes, of a label from the beginning of its enclosing segment

– Protected mode: 32 bits – Real mode: 16 bits

offset

myByte data segment:

The Protected-mode programs we write only have a single segment (we use the flat memory model).

OFFSET Examples

.data

bVal BYTE ? wVal WORD ? dVal DWORD ? dVal2 DWORD ? .code

mov esi,OFFSET bVal ; ESI = 00404000 mov esi,OFFSET wVal ; ESI = 00404001 mov esi,OFFSET dVal ; ESI = 00404003 mov esi,OFFSET dVal2; ESI = 00404007 Let's assume that bVal is located at 00404000h:

Relating to C/C++

; C++ version:

char array[1000];

char * p = &array;

The value returned by OFFSET is a pointer. Compare the following code written for both C++ and assembly language:

.data

array BYTE 1000 DUP(?) .code

mov esi,OFFSET array ; ESI is p

ALIGN Directive

• ALIGN bound aligns a variable on a byte, word, doubleword, or paragraph boundary for

efficiency. (bound can be 1, 2, 4, or 16.)

bVal BYTE ? ; 00404000 ALIGN 2

wVal WORD ? ; 00404002 bVal2 BYTE ? ; 00404004 ALIGN 4

dVal DWORD ? ; 00404008 dVal2 DWORD ? ; 0040400C

PTR Operator

.data

myDouble DWORD 12345678h .code

mov ax,myDouble ; error – why?

mov ax,WORD PTR myDouble ; loads 5678h mov WORD PTR myDouble,4321h ; saves 4321h

Overrides the default type of a label (variable).

Provides the flexibility to access part of a variable.

To understand how this works, we need to know about little endian ordering of data in memory.

Little Endian Order

• Little endian order refers to the way Intel stores integers in memory.

• Multi-byte integers are stored in reverse order, with the least significant byte stored at the lowest address

• For example, the doubleword 12345678h would be stored as:

78 0000 56 34 12

0001 0002 0003 offset byte

When integers are loaded from memory into registers, the bytes are automatically re-reversed into their correct positions.

PTR Operator Examples

.data

myDouble DWORD 12345678h

12345678 5678 0000

1234 78 56 34 12

0001 0002 0003 offset doubleword word byte

myDouble myDouble + 1 myDouble + 2 myDouble + 3

mov al,BYTE PTR myDouble ; AL = 78h mov al,BYTE PTR [myDouble+1] ; AL = 56h mov al,BYTE PTR [myDouble+2] ; AL = 34h mov ax,WORD PTR [myDouble] ; AX = 5678h mov ax,WORD PTR [myDouble+2] ; AX = 1234h

PTR Operator

.data

myBytes BYTE 12h,34h,56h,78h .code

mov ax,WORD PTR [myBytes] ; AX = 3412h mov ax,WORD PTR [myBytes+2] ; AX = 5634h mov eax,DWORD PTR myBytes ; EAX

; =78563412h PTR can also be used to combine elements of a smaller data type and move them into a larger operand. The CPU will automatically reverse the bytes.

Your turn . . .

.data

varB BYTE 65h,31h,02h,05h varW WORD 6543h,1202h varD DWORD 12345678h .code

mov ax,WORD PTR [varB+2] ; a.

mov bl,BYTE PTR varD ; b.

mov bl,BYTE PTR [varW+2] ; c.

mov ax,WORD PTR [varD+2] ; d.

mov eax,DWORD PTR varW ; e.

Write down the value of each destination operand:

0502h 78h 02h 1234h 12026543h

TYPE Operator

The TYPE operator returns the size, in bytes, of a single element of a data declaration.

.data

var1 BYTE ? var2 WORD ? var3 DWORD ? var4 QWORD ? .code

mov eax,TYPE var1 ; 1 mov eax,TYPE var2 ; 2 mov eax,TYPE var3 ; 4 mov eax,TYPE var4 ; 8

LENGTHOF Operator

.data LENGTHOF

byte1 BYTE 10,20,30 ; 3 array1 WORD 30 DUP(?),0,0 ; 32 array2 WORD 5 DUP(3 DUP(?)) ; 15 array3 DWORD 1,2,3,4 ; 4 digitStr BYTE "12345678",0 ; 9 .code

mov ecx,LENGTHOF array1 ; 32

The LENGTHOF operator counts the number of elements in a single data declaration.

SIZEOF Operator

.data SIZEOF

byte1 BYTE 10,20,30 ; 3 array1 WORD 30 DUP(?),0,0 ; 64 array2 WORD 5 DUP(3 DUP(?)) ; 30 array3 DWORD 1,2,3,4 ; 16 digitStr BYTE "12345678",0 ; 9 .code

mov ecx,SIZEOF array1 ; 64

The SIZEOF operator returns a value that is equivalent to multiplying LENGTHOF by TYPE.

Spanning Multiple Lines

.data

array WORD 10,20, 30,40,

50,60 .code

mov eax,LENGTHOF array ; 6 mov ebx,SIZEOF array ; 12

A data declaration spans multiple lines if each line (except the last) ends with a comma. The LENGTHOF and SIZEOF operators include all lines belonging to the declaration:

Spanning Multiple Lines

.data

arrayWORD 10,20 WORD 30,40 WORD 50,60 .code

mov eax,LENGTHOF array ; 2 mov ebx,SIZEOF array ; 4

In the following example, array identifies only the first WORD declaration. Compare the values returned by LENGTHOF and SIZEOF here to those in the previous slide:

LABEL Directive

• Assigns an alternate label name and type to an existing storage location

• LABEL does not allocate any storage of its own; it is just an alias.

• Removes the need for the PTR operator .data

dwList LABEL DWORD wordList LABEL WORD

intList BYTE 00h,10h,00h,20h .code

mov eax,dwList ; 20001000h mov cx,wordList ; 1000h mov dl,intList ; 00h

Indirect Addressing

• Indirect Operands

• Array Sum Example

• Indexed Operands

• Pointers

Indirect Operands

.data

val1 BYTE 10h,20h,30h .code

mov esi,OFFSET val1

mov al,[esi] ; dereference ESI (AL = 10h) inc esi

mov al,[esi] ; AL = 20h inc esi

mov al,[esi] ; AL = 30h

An indirect operand holds the address of a variable, usually an array or string. It can be dereferenced (just like a pointer). [reg] uses reg as pointer to access memory

Indirect Operands

.data

myCount WORD 0 .code

mov esi,OFFSET myCount

inc [esi] ; error: ambiguous inc WORD PTR [esi] ; ok

Use PTR when the size of a memory operand is ambiguous.

unable to determine the size from the context

Array Sum Example

.data

arrayW WORD 1000h,2000h,3000h .code

mov esi,OFFSET arrayW mov ax,[esi]

add esi,2 ; or: add esi,TYPE arrayW add ax,[esi]

add esi,2 ; increment ESI by 2 add ax,[esi] ; AX = sum of the array Indirect operands are ideal for traversing an array. Note that the register in brackets must be incremented by a value that matches the array type.

Indexed Operands

.data

arrayW WORD 1000h,2000h,3000h .code

mov esi,0

mov ax,[arrayW + esi] ; AX = 1000h mov ax,arrayW[esi] ; alternate format add esi,2

add ax,[arrayW + esi]

etc.

An indexed operand adds a constant to a register to generate an effective address. There are two notational

forms: [label + reg] label[reg]

Index Scaling

.data

arrayB BYTE 0,1,2,3,4,5 arrayW WORD 0,1,2,3,4,5 arrayD DWORD 0,1,2,3,4,5 .code

mov esi,4

mov al,arrayB[esi*TYPE arrayB] ; 04 mov bx,arrayW[esi*TYPE arrayW] ; 0004 mov edx,arrayD[esi*TYPE arrayD] ; 00000004

You can scale an indirect or indexed operand to the offset of an array element. This is done by multiplying the index by the array's TYPE:

Pointers

.data

arrayW WORD 1000h,2000h,3000h ptrW DWORD arrayW

.code

mov esi,ptrW

mov ax,[esi] ; AX = 1000h You can declare a pointer variable that contains the offset of another variable.

JMP and LOOP Instructions

• Transfer of control or branch instructions

– unconditional – conditional

• JMP Instruction

• LOOP Instruction

• LOOP Example

• Summing an Integer Array

• Copying a String

JMP Instruction

top:

. .

jmp top

• JMPis an unconditional jump to a label that is usually within the same procedure.

• Syntax: JMP target

• Logic: EIP ← target

• Example:

LOOP Instruction

• The LOOP instruction creates a counting loop

• Syntax: LOOP target

• Logic:

• ECX ← ECX – 1

• if ECX != 0, jump to target

• Implementation:

• The assembler calculates the distance, in bytes, between the current location and the offset of the target label. It is called the relative offset.

• The relative offset is added to EIP.

LOOP Example

The following loop calculates the sum of the integers 5 + 4 + 3 +2 + 1:

When LOOP is assembled, the current location = 0000000E.

Looking at the LOOP machine code, we see that –5 (FBh) is added to the current location, causing a jump to location 00000009:

00000009 ← 0000000E + FB

00000000 66 B8 0000 mov ax,0 00000004 B9 00000005 mov ecx,5 00000009 66 03 C1 L1:add ax,cx 0000000C E2 FB loop L1 0000000E

offset machine code source code

Your turn . . .

If the relative offset is encoded in a single byte, (a) what is the largest possible backward jump?

(b) what is the largest possible forward jump?

(a) −128 (b) +127

Average sizes of machine instructions are about 3 bytes, so a loop might contain, on average, a maximum of 42 instructions!

Your turn . . .

What will be the final value of AX?

mov ax,6 mov ecx,4 L1:

inc ax loop L1

How many times will the loop

execute? mov ecx,0

X2:

inc ax loop X2 10

4,294,967,296

Nested Loop

If you need to code a loop within a loop, you must save the outer loop counter's ECX value. In the following example, the outer loop executes 100 times, and the inner loop 20 times.

.data

count DWORD ? .code

mov ecx,100 ; set outer loop count L1:

mov count,ecx ; save outer loop count mov ecx,20 ; set inner loop count L2:...

loop L2 ; repeat the inner loop mov ecx,count ; restore outer loop count loop L1 ; repeat the outer loop

Summing an Integer Array

.data

intarray WORD 100h,200h,300h,400h .code

mov edi,OFFSET intarray ; address mov ecx,LENGTHOF intarray ; loop counter

mov ax,0 ; zero the sum

L1:

add ax,[edi] ; add an integer add edi,TYPE intarray ; point to next loop L1 ; repeat until ECX = 0 The following code calculates the sum of an array of 16-bit integers.

Copying a String

.data

source BYTE "This is the source string",0 target BYTE SIZEOF source DUP(0),0

.code

mov esi,0 ; index register mov ecx,SIZEOF source ; loop counter L1:

mov al,source[esi] ; get char from source mov target[esi],al ; store in the target inc esi ; move to next char loop L1 ; repeat for entire string

good use of SIZEOF

The following code copies a string from source to target.

Summary

• Data Transfer

– MOV–data transfer from source to destination – MOVSX, MOVZX, XCHG

• Operand types

– direct, direct-offset, indirect, indexed

• Arithmetic

– INC, DEC, ADD, SUB, NEG – Sign, Carry, Zero, Overflow flags

• Operators

– OFFSET, PTR, TYPE, LENGTHOF, SIZEOF, TYPEDEF

• JMP and LOOP – branching instructions

Homework #1

• Hw#1 will be returned today. Avg=94.38

0 5 10 15 20 25 30 35

50 55 60 65 70 75 80 85 90 95 99 100

Homework #3

• Assigned today, due three weeks later.

• CRC32 checksum. Refer to last year’s notes.