F28PL1 Programming Languages

Lecture 4: Assembly Language 3

Not enough registers?

• need to use memory for:– larger data structures– temporary storage of partial values

• stack– partial results e.g. during arithmetic– hold state & parameters during function call

• must allocate & manage all other memory explicitly

Stack

• stack pointer == R13 == SP• descending stack• stack pointer– decremented on PUSH– incremented on POP

– SP starts at 0x20000200 for STM32-Discovery

Stack

PUSH {Rd} SP = SP-4(*SP) = Rdi.e. • decrement SP by 4 bytes == 1 word down– to next free stack location

• store Rd at memory address indicated by SP

Stack

POP {Rd} Rd = (*SP)SP = SP+4i.e. • set Rd to contents of address indicated by SP• increment SP by 4 bytes == 1 word up– next free stack location is last used

Example: swap R1 & R2

PUSH {R1}PUSH {R2}POP {R1}POP {R2}

R1

R2

R13 0x20000200

0x200001FC

0x200001F8

0x20000200

0x22

0x11

Example: swap R1 & R2R1

R2

R13 0x20000200

0x200001FC

0x200001F8

0x200001FC

0x22

0x11

PUSH {R1}PUSH {R2}

0x11

PUSH {R1}PUSH {R2}POP {R1}POP {R2}

Example: swap R1 & R2R1

R2

R13 0x20000200

0x200001FC

0x200001F8

0x200001F8

0x22

0x11

0x11

0x22

PUSH {R1}PUSH {R2}POP {R1}POP {R2}

Example: swap R1 & R2R1

R2

R13 0x20000200

0x200001FC

0x200001F8

0x200001FC

0x22

0x22

0x11

0x22

PUSH {R1}PUSH {R2}POP {R1}POP {R2}

Example: swap R1 & R2R1

R2

R13 0x20000200

0x200001FC

0x200001F8

0x20000200

0x11

0x22

0x11

0x22

PUSH {R1}PUSH {R2}POP {R1}POP {R2}

Evaluation order

• may wish to rearrange order of evaluation• to optimise register use/number of instructionsexp1 + exp2 exp2 + exp1

e.g. A+B*C B*C+Aexp1 * exp2 exp2 * exp1

e.g. A*(B+C) (B+C)*Aexp1 * exp2 + exp1 * exp3 exp1 * (exp2 + exp3)

e.g. X*X+X*Y X*(X+Y) (X+Y)*X

Evaluation order

• sometimes need to preserve left to right order• e.g. expression contains function calls that change

shared variablesint inc(int * x){ *x = *x+1; return *x; }int a;a = 2;a+inc(&a) 2+3 5inc(&a)+a 3+3 6

Example: (a-b)/((c-d)/((e-f)/(g-h)))AREA BIGDIV,

CODE; Main__main PROC

EXPORT __main

A RN 1B RN 2C RN 3D RN 4E RN 5F RN 6G RN 7

H RN 8

MOV A,#128

MOV B,#64MOV C,#32MOV D,#16MOV E,#8MOV F,#4MOV G,#2MOV H,#1

• strict left to right• suppose only R9 & R10

spare

Example: (a-b)/((c-d)/((e-f)/(g-h)))SUB R9,A,B

• R9 == A-BSUB R10,C,D

• R10 == C-D• need register for E-F• put A-B on stack

PUSH {R9}• *SP == A-B

SUB R9,E,F• R9 == E-F• need register for G-H• put C-D on stack

PUSH {R10}• *SP == C-D

SUB R10,G,H• R10 == G-H

UDIV R9,R10• R9 == (E-F)/(G-H)• R10 free• get C-D from stack

POP {R10}• R10 == C-D

UDIV R10,R9• R10 == (C-D)/((E-F)/(G-H))

Example: (a-b)/((c-d)/((e-f)/(g-h)))• R9 free• get A-B from stack

POP {R9}• R9 == A-B

UDIV R9,R10• R9 == (A-B)/((C-D)/((E-F)/(G-H)))

ENDL B ENDLENDPEND

Areas of memory

AREA name• defines section of memory• may be manipulated as a unit by compilerAREA name,attribute1,...,attributeN

• attributeCODE == machine instructionsDATA == data

Areas of memory

READONLY == not writeable• may be in ROM• default for CODEREADWRITE == writeable• default for DATA

Allocating memory

MAP expr• set start of a storage map• starting at address expr{VAR}• assembler storage map location counter• starts at expr• e.g.MAP 0x20000000 • {VAR} == 0x20000000 == start of SRAM

Allocating memory

{label} FIELD expr• allocate expr bytes from {VAR}• label is associated with initial value of {VAR}• {VAR} = {VAR} + expr• e.g.X FIELD 4• X is 4 bytes from 0x20000000• {VAR} is now 0x20000004

Memory access

• cannot access memory directly• load register with memory address• access memory indirect on register

• load register with absolute address using MOV

LDR Rd,=label • load Rd with address corresponding to label

Memory access

[Rd] == indirection• use contents of Rd as addressLDR Rt,[Rn] Rt = *Rn• i.e. load Rt from memory whose address is in RnSTR Rt,[Rn] *Rn = Rt • i.e. store Rt at memory whose address is in Rn

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

AREA SWAP, DATA, READWRITE

SRAM EQU 0x20000000

MAP SRAM

X FIELD 4Y FIELD 4T FIELD 4 0x20000008

0x20000004

0x20000000

T

Y

X

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

AREA SWAP, CODE

; Main__main PROC

EXPORT __main

LDR R1,=XLDR R2,=YLDR R3,=T

MOV R4,#22STR R4,[R1]MOV R4,#33STR R4,[R2]

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

AREA SWAP, CODE

; Main__main PROC

EXPORT __main

LDR R1,=XLDR R2,=YLDR R3,=T

MOV R4,#22STR R4,[R1]MOV R4,#33STR R4,[R2]

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

22

22

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

AREA SWAP, CODE

; Main__main PROC

EXPORT __main

LDR R1,=XLDR R2,=YLDR R3,=T

MOV R4,#22STR R4,[R1]MOV R4,#33STR R4,[R2]

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

33

22

33

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

LDR R4,[R1]

STR R4,[R3]

LDR R4,[R2]

STR R4,[R1]

LDR R4,[R3]

STR R4,[R2]

ENDL B ENDLENDP

END

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

22

22

33

22

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

LDR R4,[R1]

STR R4,[R3]

LDR R4,[R2]

STR R4,[R1]

LDR R4,[R3]

STR R4,[R2]

ENDL B ENDLENDP

END

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

22

22

33

22

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

LDR R4,[R1]

STR R4,[R3]

LDR R4,[R2]

STR R4,[R1]

LDR R4,[R3]

STR R4,[R2]

ENDL B ENDLENDP

END

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

33

22

33

22

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

LDR R4,[R1]

STR R4,[R3]

LDR R4,[R2]

STR R4,[R1]

LDR R4,[R3]

STR R4,[R2]

ENDL B ENDLENDP

END

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

33

33

33

22

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

LDR R4,[R1]

STR R4,[R3]

LDR R4,[R2]

STR R4,[R1]

LDR R4,[R3]

STR R4,[R2]

ENDL B ENDLENDP

END

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

22

33

33

22

Example: swap a & bint x;int y;int t;x = 22;y = 33;t = x;x = y;y = t;

LDR R4,[R1]

STR R4,[R3]

LDR R4,[R2]

STR R4,[R1]

LDR R4,[R3]

STR R4,[R2]

ENDL B ENDLENDP

END

R1

R2

R30x20000008

0x20000004

0x20000000

R4

0x20000008

0x20000004

0x20000000

T

Y

X

22

33

22

22

Array space

• array is continuous sequence of bytes

type name[size]• allocate sizeof(type)*size bytes• start register at address of 1st byte• access byte/half word/word indirect on register• add 1/2/4 to register to get to address for next

byte/halfword/word

Array access

• a[i] == address for a + i*size(type)

• for iterative access to a[i]– 0<= i<size

• put address of array in register Ri• on each iteration– access [Ri]– add size(type) to Ri

Example: a[i] = i#define MAX 25int a[MAX];int i;i = 0;while(i<MAX){ a[i] = i; i = i+1;}

AREA SETA, DATA, READWRITE

SRAM EQU 0x20000000MAP SRAM

MAX EQU 25A FIELD MAX*4 • A is 0x20000000 to 0x20000064

AREA SETA, CODE__main PROC

EXPORT __main

Example: a[i] = i#define MAX 25int A[MAX];int i;i = 0;while(i<MAX){ A[i] = i; i = i+1;}

LDR R1,=A - R1 == A@0x20000000MOV R2,#0 - R2 == i

TEST CMP R2,#MAXBEQ ENDLSTR R2,[R1] - (*A) = iADD R2,#1 - i = i+1ADD R1,#0x04 - A = A+4B TEST

ENDL B ENDLENDPEND

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD

R1,#0x04B TEST

ENDL B ENDLENDPEND

R1

R20

0x20000000

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R20

0x20000000

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

0

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R21

0x20000000

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

0

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R21

0x20000004

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

0

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R21

0x20000004

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

1

0

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R22

0x20000004

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

1

0

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R22

0x20000008

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

1

0

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R22

0x20000008

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

1

0

2

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R23

0x20000008

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

1

0

2

Example: a[i] = iLDR R1,=AMOV R2,#0

TEST CMP R2,#MAX

BEQ ENDLSTR R2,

[R1]ADD R2,#1ADD R1,#4B TEST

ENDL B ENDLENDPEND

R1

R23

0x2000000C

0x20000008

0x20000004

0x20000000 a[0]

a[1]

a[2]

1

0

2