pointers and arrays (part 2)

Faculty of Computer Science

CMPUT 229 © 2006

Pointers and Arrays(part 2)

Differentiating Pointers from Data

© 2006

Department of Computing Science

CMPUT 229

The car.c program#include <stdio.h>#define STRINGLENGTH 20

typedef struct c_node{ int vehicleID; char make[STRINGLENGTH]; char model[STRINGLENGTH]; int year; int mileage; double cost; struct c_node *next;} CarNode;

void ReadCar(CarNode *car);void PrintCar(CarNode car);

main(){ CarNode mycar; ReadCar(&mycar); PrintCar(mycar);}

void ReadCar(CarNode *car){ car->vehicleID = 2; strcpy(car->make,"DODGE"); strcpy(car->model,"STRATUS"); car->year = 1996; car->mileage = 70000; car->cost = 4,525.74;}

void PrintCar(CarNode car){ printf("vehicleID: %d\n",car.vehicleID); printf("make: %s\n",car.make); printf("model: %s\n",car.model); printf("year: %d\n",car.year); printf("mileage: %d\n",car.mileage); printf("cost: %f\n",car.cost);}

Patt and Patel, pp. 419

© 2006


CMPUT 229

The car.c program

#define STRINGLENGTH 20

typedef struct c_node{ int vehicleID; char make[STRINGLENGTH]; char model[STRINGLENGTH]; int year; int mileage; double cost; struct c_node next;} CarNode;

vehicleID

make[20]

model[20]

yearmileage

cost

next

048

12162024283236404448525660

© 2006


CMPUT 229

The car.c programmain:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

#include <stdio.h>#define STRINGLENGTH 20

typedef struct c_node{ int vehicleID; char make[STRINGLENGTH]; char model[STRINGLENGTH]; int year; int mileage; double cost; struct c_node *next;} CarNode;

void ReadCar(CarNode *car);void PrintCar(CarNode car);

main(){ CarNode mycar; ReadCar(&mycar); PrintCar(mycar);}

© 2006


CMPUT 229

The car.c program

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SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

© 2006


CMPUT 229

The car.c program

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92SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

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A6<old A6>

© 2006


CMPUT 229

The car.c program

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92SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

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A6<old A6>

© 2006


CMPUT 229

The car.c programmake[0-3]

vehicleID

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92SP

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

© 2006


CMPUT 229


vehicleID

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92SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

© 2006


CMPUT 229


vehicleID

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SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

© 2006


CMPUT 229


vehicleID

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92

SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

D024

D124

A088

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

© 2006


CMPUT 229


vehicleID

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SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

D064

D124

A088

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

© 2006


CMPUT 229

The car.c program

648824

make[0-3]

vehicleID

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SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

D064

D124

A088

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

© 2006


CMPUT 229

memcpyNAME

memcpy - copy memory area

SYNOPSIS

#include <string.h>

void *memcpy(void *dest, const void *src, size_t n);

DESCRIPTION

The memcpy() function copies n bytes from memory area src to memory

area dest. The memory areas should not overlap. Use memmove(3) if the

memory areas do overlap.

RETURN VALUE

The memcpy() function returns a pointer to dest.

© 2006


CMPUT 229

The car.c program

648824

make[0-3]

vehicleID

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SP

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

addq.l #4,SP

lea (-64,SP),SP

move.l SP,D0

move.l D0,D1

lea (-68,A6),A0

moveq #64,D0

move.l D0,-(SP)

move.l A0,-(SP)

move.l D1,-(SP)

jbsr memcpy

lea (12,SP),SP

jbsr PrintCar

lea (64,SP),SP

unlk A6

rts

D064

D124

A088

<old A6>next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

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A6

cost2

cost1

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

make[0-3]

vehicleID

next

© 2006


CMPUT 229

Why so much copying?

The program car.c passes the data structure CarNodeto the PrintCar function by value.

A copy of each byte of CarNode must be made in thestack for each call of the function PrintCar.

We could, instead have passed the address of the copyof CarNode that we already had in the stack.

© 2006


CMPUT 229

The car2.c Program#include <stdio.h>#define STRINGLENGTH 20


void ReadCar(CarNode car);void PrintCar(CarNode car);

main(){ CarNode mycar; ReadCar(&mycar); PrintCar(&mycar);}

void ReadCar(CarNode car){ car->vehicleID = 2; strcpy(car->make,"DODGE"); strcpy(car->model,"STRATUS"); car->year = 1996; car->mileage = 70000; car->cost = 4,525.74;}

void PrintCar(CarNode car){ printf("vehicleID: %d\n",car->vehicleID); printf("make: %s\n",car->make); printf("model: %s\n",car->model); printf("year: %d\n",car->year); printf("mileage: %d\n",car->mileage); printf("cost: %f\n",car->cost);}

© 2006


CMPUT 229

Assembly for car2.c

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

jbsr PrintCar

addq.l #4,SP

unlk A6

rts

#include <stdio.h>#define STRINGLENGTH 20


void ReadCar(CarNode car);void PrintCar(CarNode car);

main(){ CarNode mycar; ReadCar(&mycar); PrintCar(&mycar);}

© 2006


CMPUT 229

Assembly for car2.c

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

jbsr PrintCar

addq.l #4,SP

unlk A6

rts

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92SP

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A6<old A6>

© 2006


CMPUT 229

Assembly for car2.c

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

jbsr PrintCar

addq.l #4,SP

unlk A6

rts

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92SP

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A6<old A6>

A088

© 2006


CMPUT 229

Assembly for car2.c

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

jbsr PrintCar

addq.l #4,SP

unlk A6

rts

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92SP

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A6<old A6>

A088

© 2006


CMPUT 229

Assembly for car2.c

main:

link.w A6,#-68

lea (-68,A6),A0

move.l A0,-(SP)

jbsr ReadCar

jbsr PrintCar

addq.l #4,SP

unlk A6

rts

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92SP

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A6<old A6>

A088

next

cost

cost

mileage

year

model[16-19]

model[12-15]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

make[0-3]

vehicleID

© 2006


CMPUT 229

Quiz #1

predecessors and successors are two vectors of 32-bit unsigned integers.Assume that:

predecessors’ first element is stored at A6+32successors’ first element is stored at A6+64

Where A6 points to the base of the stack frame of the current procedure.

Write a sequence of assembly instructions that executes the followingC statement:

successors[5] = predecessors[7];

© 2006


CMPUT 229

Quiz #1 - Solution

predecessors and successors are two vectors of 32-bit unsigned integers.predecessors’ first element is stored at A6+32successors’ first element is stored at A6+64


D0 predecessors[7];successors[5] D0;

A1 Addr(predecessor[7]);D0 (A1);A2 Addr(successors[5]);(A2) D0;

A1 A6+32+7*4D0 (A1);A2 A6+64+5*4(A2) D0;

move.l A6, A1 addq.l #60,A1

move.l (A1), D0 move.l A6, A2

addq.l #84, A2 move.l D0, (A2)

A1 A6+60D0 (A1);A2 A6+84(A2) D0;

Can you do better than this?

© 2006


CMPUT 229

Quiz #1 - Solution





A1 A6+32+7*4D0 0(A1);A2 A6+64+5*4(A2) D0;

move.l (A6,60), D0 move.l D0, (A6,84)

A1 A6+60D0 (A1);A2 A6+84(A2) D0;

© 2006


CMPUT 229

Quiz #1 - Solution





A1 A6+32+7*4D0 0(A1);A2 A6+64+5*4(A2) D0;

move.l (A6,60), (A6,84)

A1 A6+60D0 (A1);A2 A6+84(A2) D0;

© 2006


CMPUT 229

Quiz #2

names is an array of pointers to strings.

Each position of names contains a pointer to a null terminated string.

Assume that the first element of names is stored in the memory position SP+16.

Write a sequence of assembly instructionsthat counts the number of characters in the string

whose pointer is at names[7].

© 2006


CMPUT 229

Quiz #2

names is an array of pointers to strings. Each position of names contains a pointer to a null terminated string.Assume that the first element of names is stored in the memory position SP+16.Write a sequence of assembly instructions that counts the number of characters in the string whose pointer is at names[7].

count = 0;c = names[7];while (*c != 0) { count = count+1; c = c+1; }

C

D0 0;A2 names[7];while ( *(A2) != 0) { D0 D0+1; A2 A2+1; }

C

D0 0;A1 Address(names[7]);A2 (A1);while ((A2) != 0) { D0 D0+1; A2 A2+1; }

RTL

D0 0;A1 SP+16+7*4A2 (A1);while ((A2)+ != 0) { D0 D0+1; }

RTL

© 2006


CMPUT 229

Quiz #2

names is an array of pointers to strings. Each position of names contains a pointer to a null terminated string.Assume that the first element of names is stored in the memory position SP+16.Write a sequence of MIPS assembly instructions that counts the number of characters in the string whose pointer is at names[7].

D0 0;A1 SP+44;A2 (A1);while ((A2)+ != 0) { D0 = D0+1; }

clr D0move (SP,44), A2cmp.b (A2)+, #0beq doneaddq.l #1, D0bra loop;

done:

loop:

RTL

D0 0;A2 (SP+44);while ((A2)+ != 0) { D0 = D0+1; }

RTL

Assembly

© 2006


CMPUT 229

Quiz #2names is an array of pointers to strings. Each position of names contains a pointer to a null terminated string.Assume that the first element of names is stored in the memory position SP+16.Write a sequence of MIPS assembly instructions that counts the number of characters in the string whose pointer is at names[7].

clr D0move (SP,44), A2cmp.b (A2)+, #0beq doneaddq.l #1, D0bra loop;

loop:

Names[7]

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SP

c 2 2 9 \0 A2

D0 0;A2 (SP+44);while ((A2)+ != 0) { D0 = D0+1; }

RTL

© 2006


CMPUT 229

Quiz #3

Consider the following C function:

Write assembly code for the first statement of this function.Assume: temp D0 node D1 parameters are passed in the stack

int ReadGraph(FILE input_file, unsigned int successors, char ***names){ unsigned int temp = 0; unsigned int node = 0;

(*successors)[node] = temp; printf(“name = %d\n”,*names[node]);}

Input_filesuccessors

names<ret adr>16

20

24

28

32

36

SP

8

© 2006


CMPUT 229

The meaning of (*successor)[node]

int *vector;Int position;Int value;

vector[position] = value;

(*successor)[node] = temp;

successor

vector

int *vector;Int position;Int value;

*(vector+position) = value;

=

*((*successor)+node) = temp;=

© 2006


CMPUT 229

Quiz #3 - Solution

int ReadGraph(FILE input_file, unisgned int successors, char ***names){ unsigned int temp = 0; unsigned int node;


(*successors)[node] temp;

(SP+8)[D1] D0;

A1 (SP+8)(A1+D1*4) D0;

move.l (SP,8), A1


names<ret adr>16

20

24

28

32

36

SP

node

8

temp D0node D1

A1

© 2006


CMPUT 229

Quiz #3 - Solution

int ReadGraph(FILE input_file, unisgned int successors, char ***names){ unsigned int temp = 0; unsigned int node;


(*successors)[node] temp;

(SP+8)[D1] D0;

A1 (SP+8)(A1+D1*4) D0;

move.l (SP,8), A1 move.l D1, D2 lsl.l #2, D2

move.l (A1,D2), D0


names<ret adr>16

20

24

28

32

36

SP

node

8

temp D0node D1

A1

D0

© 2006


CMPUT 229

Quiz #4

Consider the following C function:

Assume that ReadGraph receives its parameters in the stack.Write assembly code to push the parameter (*names)[node] into the stack before the call to the function call to printf.

int ReadGraph(FILE input_file, unisgned int successors, char ***names){ unsigned int temp; unsigned int node;

(*successors)[node] = temp; printf(“name = %s\n”,(*names)[node]);}

© 2006


CMPUT 229

int ReadGraph(FILE input_file, unisgned int successors, char ***names){ printf(“”,(*names)[node]);}

SP+4

move.l (SP,4), A0 # A0 names

A0

A d a m \0

J a m e \0s

C e c i \0l y

M u r r \0a y


names<ret adr>16

20

24

28

32

36

SP

8

temp D0node D1

© 2006


CMPUT 229

int ReadGraph(FILE input_file, unisgned int successors, char ***names){ printf(“”,(*names)[node]);}

SP+4

A1

move.l (SP,4), A0 # A0 names Move.l D1, D2 # D2 4nodelsl.l #2, D1 # D1 4nodemove.l (A0,D1), A1 # A1 (names)[node] move.l A1, -(SP) # Push A1 into stack

A0

A d a m \0

J a m e \0s

C e c i \0l y

M u r r \0a y


names<ret adr>16

20

24

28

32

36

SP

8

temp D0node D1

© 2006


CMPUT 229

Quiz #7

Write assembly for the following program.Assume that:

apple A6+12 ptr A6+8ind A6+4

# 7 ind = &ptr;lea (A6,8), A0 A0 &ptr move.l A0, (A6,4) ind &ptr# 8 *ind = &apple;lea (A6,12), A0 A0 &applemove.l (A6,4), A1 A1 indmove.l A0, (A1) *ind &apple# 9 **ind = 123;move.l (A6,4), A0 A0 indmove.l (A0), A1 A1 *indmove.l #123, (A1) **ind 123# 10 ind++;addq.l #4, (A6,4) ind++# 11 (*ptr)++;move.l (A6,8), A0 A0 ptraddq.l #4, (A0) *ptr *ptr + 4# 12 apple++;Addq.l #4, (A6,12) apple apple + 4

1 #include <stdio.h> 2 main() 3 { 4 int apple; 5 int *ptr; 6 int **ind;

7 ind = &ptr; 8 *ind = &apple; 9 **ind = 123;10 ind++;11 (*ptr)++;12 apple++;

13 printf(“%x %x %d\n”, ind, ptr, apple);14 }

ind

Stackapple

ptr

A6 0

4

8

12A0

A1

© 2006


CMPUT 229

Quiz #7



# 7 ind = &ptr;lea (A6,8), A0 A0 &ptr move.l A0, (A6,4) ind &ptr# 8 *ind = &apple;lea (A6,12), A0 A0 &applemove.l (A6,4), A1 A1 indmove.l A0, (A1) *ind &apple# 9 **ind = 123;move.l (A6,4), A0 A0 indmove.l (A0), A1 A1 *indmove.l #123, (A1) **ind 123# 10 ind++;addq.l #4, (A6,4) ind++# 11 (*ptr)++;move.l (A6,8), A0 A0 ptraddq.l #4, (A0) *ptr *ptr + 4# 12 apple++;qddq.l #4, (A6,12) apple apple + 4




A6+12

Stack131

A6+12

A6 0

4

8

12A6+12

A6+12

A0

A1

pointers and arrays (part 2)

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