cmput229 - fall 2003

CMPUT 229 - Computer Organization and Architecture I

1

CMPUT229 - Fall 2003

TopicC: Pointers and Arrays

José Nelson Amaral


2

Reading Material

The slides for this topic were prepared based on chapters 17 of:

Patt, Yale N., and Patel, Sanjay J., Introduction to Computing Systems: from bits & gates to C & Beyond, McGrawHill Press, 2001.

An excellent reference book for the C Language is:

Harbison, Samuel P., and Steele Jr., Guy, C: A Reference Manual, Prentice Hall, 4th Edition, 1995.


3

Example: A swap function

1 #include <stdio.h> 2 void Swap(int firstVal, int secondVal); 3 4 main() 5 { 6 int valueA = 3; 7 int valueB = 4; 8 9 printf("Before Swap: valueA = %d and valueB = %d\n", valueA, valueB);10 Swap(valueA, valueB);11 printf("After Swap: valueA = %d and valueB = %d\n", valueA, valueB);12 }1314 void Swap(int firstVal, int secondVal)15 {16 int tempVal;1718 tempVal = firstVal;19 firstVal = secondVal;20 secondVal = tempVal;21 }

Patt and Patel, pp. 366


4

Assembly for Swap Generated with -O0 (part 1)


# 5 main() # 6 {

addiu $sp,$sp,-32 # .frame.len.mainsd $gp,16($sp) # .lcl_spill_b002sd $ra,8($sp) # .lcl_spill_b001lui $a3,%hi(%neg(%gp_rel(main +0)))addiu $a3,$a3,%lo(%neg(%gp_rel(main +0))) addu $gp,$t9,$a3

# 7 int valueA = 3;addiu $a2,$zero,3 sw $a2,0($sp) # valueA

# 8 int valueB = 4;addiu $a1,$zero,4 sw $a1,4($sp) # valueB

# 10 printf("Before Swap: valueA = %d and valueB = %d\n", valueA, valueB);lw $a0,%got_page(.rodata)($gp) addiu $a0,$a0,%got_ofst(.rodata) lw $a1,0($sp) # valueAlw $a2,4($sp) # valueBlw $t9,%call16(printf)($gp) jalr $t9 # printfnop

$a1

Stack

$a2

$sp$a0

0

4

3

434bash-2.01$ ./swapO0

Before Swap: valueA = 3 and valueB = 4


5



.BB2.main: # 0x44 # 11 Swap(valueA, valueB);

lw $a0,0($sp) # valueAlw $a1,4($sp) # valueBlw $t9,%call16(Swap)($gp) jalr $t9 # Swapnop

.BB3.main: # 0x58 # 12 printf("After Swap: valueA = %d and valueB = %d\n", valueA, valueB);

lw $a0,%got_page(.rodata+48)($gp)addiu $a0,$a0,%got_ofst(.rodata+48)lw $a1,0($sp) # valueAlw $a2,4($sp) # valueBlw $t9,%call16(printf)($gp) jalr $t9 # printfnop

.BB4.main: # 0x74 # 13 }

or $v0,$zero,$zero ld $gp,16($sp) # .lcl_spill_b002ld $ra,8($sp) # .lcl_spill_b001addiu $sp,$sp,32 # .frame.len.mainjr $ra nop

4$a1

Stack

3$a2

43$sp

$a0 30

4

bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4


6



# 15 void Swap(int firstVal, int secondVal) # 16 {

addiu $sp,$sp,-32 # .frame.len.Swapsw $a0,20($sp) # firstValsw $a1,28($sp) # secondVal

# 17 int tempVal; # 18 # 19 tempVal = firstVal;

lw $v1,20($sp) # firstValsw $v1,0($sp) # tempVal

# 20 firstVal = secondVal;lw $v0,28($sp) # secondValsw $v0,20($sp) # firstVal

# 21 secondVal = tempVal;lw $at,0($sp) # tempValsw $at,28($sp) # secondVal

# 22 }addiu $sp,$sp,32 # .frame.len.Swapjr $ra nop .end Swap

4$a1

3$a2

3$a0

Stack43$sp 0

4



7










4$a1

3$a2

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Stack43

$sp

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$v0

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3

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$at 3

3

0

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4$a1

3$a2

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Stack43$sp

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4$v0

$at 3

0

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9






.BB4.main: # 0x74 # 13 }



4$a1

3$a2

3$a0

4

3

3

Stack43$sp

3$v1

4$v0

$at 3

0

4

bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4


10






.BB4.main: # 0x74 # 13 }



3$a1

4$a2

3$a0

4

3

3

Stack43$sp

3$v1

4$v0

$at 3

0

4

bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4After Swap: valueA = 3 and valueB = 4


11

Addresses and Values

The problem with our swap program is that the main is passingthe values of variables valueA and valueB to the swap function.

Thus the swap function does all its work within its own frame inthe stack and never actually changes the state of the variablesin the main function.

Could a “smarter” compiler figure out that the swap function is doing nothing? Lets try the MIPSPro compiler with -O3.


12



# 5 main() # 6 {

lui $a1,%hi(%neg(%gp_rel(main +0))) # [0] addiu $sp,$sp,-16 # [0] .frame.len.mainsd $gp,8($sp) # [1] .gra_spill_b002addiu $a1,$a1,%lo(%neg(%gp_rel(main +0))) # [1] addu $gp,$t9,$a1 # [2] lw $t9,%got_disp(printf)($gp) # [3] .loc 1 10 3

# 7 int valueA = 3; # 8 int valueB = 4; # 10 printf("Before Swap: valueA = %d and valueB = %d\n", valueA, valueB);

lw $a0,%got_page(.rodata)($gp) # [4] addiu $a2,$zero,4 # [5] sd $ra,0($sp) # [5] .gra_spill_b001addiu $a1,$zero,3 # [5] jalr $t9 # [6] printfaddiu $a0,$a0,%got_ofst(.rodata) # [6]

.BB2.main: # 0x30 #<freq> BB:2 frequency = 1.00000 (heuristic)# 11 Swap(valueA, valueB);

lw $t9,%call16(Swap)($gp) # [0] addiu $a1,$zero,4 # [2] jalr $t9 # [3] Swapaddiu $a0,$zero,3 # [3]

$a1

Stack

$a2

$sp$a0

0

4

4

3



13



# 5 main() # 6 {

lui $a1,%hi(%neg(%gp_rel(main +0))) # [0] addiu $sp,$sp,-16 # [0] .frame.len.mainsd $gp,8($sp) # [1] .gra_spill_b002addiu $a1,$a1,%lo(%neg(%gp_rel(main +0))) # [1] addu $gp,$t9,$a1 # [2] lw $t9,%got_disp(printf)($gp) # [3] .loc 1 10 3

# 7 int valueA = 3; # 8 int valueB = 4; # 10 printf("Before Swap: valueA = %d and valueB = %d\n", valueA, valueB);

lw $a0,%got_page(.rodata)($gp) # [4] addiu $a2,$zero,4 # [5] sd $ra,0($sp) # [5] .gra_spill_b001addiu $a1,$zero,3 # [5] jalr $t9 # [6] printfaddiu $a0,$a0,%got_ofst(.rodata) # [6]

.BB2.main: # 0x30 #<freq> BB:2 frequency = 1.00000 (heuristic)# 11 Swap(valueA, valueB);

lw $t9,%call16(Swap)($gp) # [0] addiu $a1,$zero,4 # [2] jalr $t9 # [3] Swapaddiu $a0,$zero,3 # [3]

$a1

Stack

$a2

$sp$a0

0

4

4

3



14



Swap: # 0x70.frame $sp, 0, $ra

.BB1.Swap: # 0x70 #<freq> #<freq> BB:1 frequency = 1.00000 (heuristic) #<freq> # 18 # 19 tempVal = firstVal; # 20 firstVal = secondVal; # 21 secondVal = tempVal; # 22 }

jr $ra # [0] nop # [0]

$a1

Stack

$a2

$sp$a0

0

4

4

3



15



.BB3.main: # 0x40 #<freq> #<freq> BB:3 frequency = 1.00000 (heuristic) #<freq>

lw $t9,%got_disp(printf)($gp) # [0] # 12 printf("After Swap: valueA = %d and valueB = %d\n", valueA, valueB);

lw $a0,%got_page(.rodata+48)($gp) # [1] addiu $a2,$zero,4 # [2] addiu $a1,$zero,3 # [2] jalr $t9 # [3] printfaddiu $a0,$a0,%got_ofst(.rodata+48) # [3]

.BB4.main: # 0x58 #<freq> BB:4 frequency = 1.00000 (heuristic) # 13 }

ld $ra,0($sp) # [0] .gra_spill_b001or $v0,$zero,$zero # [2] ld $gp,8($sp) # [3] .gra_spill_b002jr $ra # [3] addiu $sp,$sp,16 # [3] .frame.len.main

$a1

Stack

$a2

$sp$a0

0

4

3

4

bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4bash-2.01$ ./swapO0Before Swap: valueA = 3 and valueB = 4After Swap: valueA = 3 and valueB = 4


16

What happened at -O3?

While compiling the swap code, MIPSPro figured out that it wasdoing nothing of consequence, and thus replaced the body of thefunction with a simple return instruction.

However during the compilation of main, MIPSPro did not knowwhat swap was up to, and thus could not eliminate the call toswap.

This happens because, even at -O3, MIPSPro does not do inter-procedural analysis (IPA), and thus the compilation ofeach procedure is done in isolation.

In order to force MIPSPro do perform IPA, we would have toinclude the option -ipa in the command line. But then the compiler cannot produce the assembly files because IPA actuallytakes place during the linking phase of the compiler.


17

Example2: A new swap function

1 #include <stdio.h> 2 void NewSwap(int *firstVal, int *secondVal); 3 4 main() 5 { 6 int valueA = 3; 7 int valueB = 4; 8 9 printf("Before Swap: valueA = %d and valueB = %d\n", valueA, valueB);10 NewSwap(&valueA, &valueB);11 printf("After Swap: valueA = %d and valueB = %d\n", valueA, valueB);12 }1314 void NewSwap(int *firstVal, int *secondVal)15 {16 int tempVal;1718 tempVal = *firstVal;19 *firstVal = *secondVal;20 *secondVal = tempVal;21 }



18


Assembly for NewSwap Generated with -O3 (part 1) # 5 main()

# 6 {lui $a2,%hi(%neg(%gp_rel(main +0))) # [0] addiu $sp,$sp,-32 # [0] .frame.len.mainsd $gp,16($sp) # [1] .gra_spill_b002addiu $a2,$a2,%lo(%neg(%gp_rel(main +0))) # [1]

# 7 int valueA = 3;addiu $a1,$zero,3 # [2] sw $a1,0($sp) # [2] valueAaddu $gp,$t9,$a2 # [2] lw $t9,%got_disp(printf)($gp) # [3]

# 8 int valueB = 4;addiu $a3,$zero,4 # [4]

# 10 printf("Before Swap: valueA = %d and valueB = %d\n", valueA, valueB);lw $a0,%got_page(.rodata)($gp) # [4] sd $ra,8($sp) # [5] .gra_spill_b001addiu $a2,$zero,4 # [5] addiu $a1,$zero,3 # [5] sw $a3,4($sp) # [6] valueBjalr $t9 # [6] printfaddiu $a0,$a0,%got_ofst(.rodata) # [6]

#<freq> BB:2 frequency = 1.00000 (heuristic) # 11 NewSwap(&valueA, &valueB);

lw $t9,%call16(NewSwap)($gp) # [0] addiu $a1,$sp,4 # [2] valueBjalr $t9 # [3] NewSwapaddiu $a0,$sp,0 # [3] valueA

Stack

0

4

$a2

$a3

$a0

$a1 33

4

0x7ffff0004

40x7ffff0000bash-2.01$ ./newswapO0Before Swap: valueA = 3 and valueB = 4 $sp

4


19


Assembly for NewSwap Generated with -O3 (part 1)

# 15 void NewSwap(int *firstVal, int *secondVal) # 16 { # 17 int tempVal; # 18 # 19 tempVal = *firstVal;

lw $at,0($a0) # 20 *firstVal = *secondVal;

lw $v0,0($a1) sw $v0,0($a0)

# 21 *secondVal = tempVal; # 22 }

jr $ra sw $at,0($a1)

Stack43 0

4

$a2

$a3 4

$a0 0x7ffff0000

$a1 0x7ffff0004

$at

$v0

$sp

3

34

4

bash-2.01$ ./newswapO0Before Swap: valueA = 3 and valueB = 4


20

Assembly for NewSwap Generated with -O3 (part 1)


lw $t9,%got_disp(printf)($gp)

# 12 printf("After Swap: valueA = %d and valueB = %d\n", valueA, valueB);lw $a0,%got_page(.rodata+48)($gp)lw $a2,4($sp) # [2] valueBlw $a1,0($sp) # [3] valueAjalr $t9 # [3] printfaddiu $a0,$a0,%got_ofst(.rodata+48)

Stack34 0

4

$a2

$a3 4

$a0 0x7ffff0000

$a1 0x7ffff0004

$at 3

$v0 4

$sp

3

4

bash-2.01$ ./newswapO0Before Swap: valueA = 3 and valueB = 4bash-2.01$ ./newswapO0Before Swap: valueA = 3 and valueB = 4After Swap: valueA = 4 and valueB = 3


21

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);}



22

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

1216202428323640444852566064


23

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


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

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

main:subu $sp, 168sw $ra, 84($sp).frame $sp, 168, $ra

# 23 CarNode mycar; # 25 ReadCar(&mycar);

addu $a0, $sp, 88jal ReadCar

# 27 PrintCar(mycar);addu $t6, $sp, 88move $t9, $spaddu $t0, $t6, 72

L1:lw $t8, 0($t6)addu $t6, $t6, 12sw $t8, 0($t9)lw $t7, -8($t6)addu $t9, $t9, 12sw $t7, -8($t9)lw $t8, -4($t6)sw $t8, -4($t9)bne $t6, $t0, L1lw $a0, 0($sp)lw $a1, 4($sp)lw $a2, 8($sp)lw $a3, 12($sp)jal PrintCar

# 28 }move $v0, $zerolw $ra, 84($sp)addu $sp, 168j $ra


24

The car.c program







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

SP00

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$t9

$t6

$t0

$t8

$t7

vehicleID


25

The car.c program







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

SP00

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$t9

$t0

$t8

$t7

vehicleID

$t6

vehicleID

make[0-3]


26

The car.c program







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

SP00

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$t0

$t8

$t7

vehicleID

$t6

vehicleID

make[0-3]

$t9

make[0-3]

make[4-7]

make[4-7]


27

The car.c program







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

SP00

04

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$t0

$t8

$t7

$t6

vehicleID

make[0-3]

$t9

make[0-3]

make[4-7]

make[4-7]

make[8-11]


28

The car.c program







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

SP00

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$t0

$t8

$t7

vehicleID

make[0-3]

$t9

make[0-3]

make[4-7]

make[8-11]

$t6

make[8-11]

make[12-15]


29

The car.c program







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[8-11]

make[4-7]

make[0-3]

make[0-3]

vehicleID

SP00

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$t0

$t8

$t7

make[8-11]

vehicleID

make[12-15]

$t9

make[12-15]

make[16-19]

make[16-19]$t6


30

The car.c program







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[16-19]

make[12-15]

make[8-11]

make[0-3]

vehicleID

SP00

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$t0

$t8

$t7

vehicleID

make[12-15]

make[0-3]

make[16-19]

make[4-7]

model[0-3]

$t6

$t9


31

The car.c program







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[16-19]

make[12-15]

make[8-11]

make[0-3]

vehicleID

SP00

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$t0

$t8

$t7

vehicleID

make[12-15]

make[0-3]

make[4-7]

model[0-3]

$t6

model[0-3]

model[4-7]

$t9


32

The car.c program







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]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[4-7]

make[0-3]

make[0-3]

vehicleID

SP00

04

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$t0

$t8

$t7

model[0-3]

vehicleID

model[4-7]

$t9

model[4-7]

model[8-11]

model[8-11]$t6


33

The car.c program







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]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

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68

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96

100

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108

112116

120

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136

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144148

152

156

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176

$t0

$t8

$t7

vehicleID

model[4-7]

make[0-3]

model[8-11]

make[4-7]

model[12-15]

$t6

$t9


34

The car.c program







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]

model[8-11]

model[4-7]

model[0-3]

make[16-19]

make[12-15]

make[8-11]

make[0-3]

vehicleID

SP00

04

08

12

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20

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32

36

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44

48

52

56

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68

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96

100

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108

112116

120

124

128

132

136

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144148

152

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164

168

172

176

$t0

$t8

$t7

vehicleID

make[12-15]

make[0-3]

make[4-7]

model[12-15]

$t6

model[12-15]

model[16-19]

$t9


35

The car.c program







next

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]

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]

make[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

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44

48

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56

60

64

68

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188

96

100

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112116

120

124

128

132

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144148

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156

160

164

168

172

176

$t0

$t8

$t7

model[12-15]

vehicleID

model[16-19]

$t9

model[16-19]

year

year$t6


36

The car.c program







next

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]

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[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

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92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

$t8

$t7

vehicleID

model[16-19]

make[0-3]

year

make[4-7]

mileage

$t6

$t9


37

The car.c program







next

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]

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[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

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92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

$t8

$t7

vehicleID

make[16-19]

make[0-3]

make[4-7]

mileage

$t6

mileage

$t9


38

The car.c program







next

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]

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]

make[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

$t8

$t7

model[12-15]

vehicleID

$t9

cost1

cost1$t6


39

The car.c program







next

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]

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[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

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92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

vehicleID

make[0-3]

make[4-7]

$t8

$t7

$t9

cost1

$t6

cost2


40

The car.c program







next

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]

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[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

vehicleID

make[0-3]

make[4-7]

$t6

cost2

$t8

$t7

$t9

cost2

next


41

The car.c program







next

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]

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]

make[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

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92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

$t8

$t7

cost2

vehicleID

next

next

$t9$t6


42

The car.c program







next

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]

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]

make[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

$t0

$t8

$t7

cost2]

vehicleID

next

$t9

next

$t6

$a0vehicleID

$a1make[0-3]

$a2make[4-7]

$a3make[8-11]


43

The car.c program # 42 void PrintCar(CarNode car) # 43 {PrintCar:

subu $sp, 32sw $ra, 28($sp)sw $a0, 32($sp)sw $a1, 36($sp)sw $a2, 40($sp)sw $a3, 44($sp)

# 44 printf("vehicleID: %d\n",car.vehicleID);la $a0, string13lw $a1, 32($sp)jal printf

# 45 printf("make: %s\n",car.make);la $a0, string14addu $a1, $sp, 32addu $a1, $a1, 4jal printf

# 46 printf("model: %s\n",car.model);la $a0, string15addu $a1, $sp, 32addu $a1, $a1, 24jal printf

# 47 printf("year: %d\n",car.year);la $a0, string16lw $a1, 76($sp)jal printf

next

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]

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]

make[0-3]

vehicleID

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

180

184

188

96

100

104

108

112116

120

124

128

132

136

140

144148

152

156

160

164

168

172

176

vehicleID

next

$a0vehicleID

$a1make[0-3]

$a2make[4-7]

$a3make[8-11]


44

The car.c program # 42 void PrintCar(CarNode car) # 43 {PrintCar:

subu $sp, 32sw $ra, 28($sp)sw $a0, 32($sp)sw $a1, 36($sp)sw $a2, 40($sp)sw $a3, 44($sp)

# 44 printf("vehicleID: %d\n",car.vehicleID);la $a0, string13lw $a1, 32($sp)jal printf

# 45 printf("make: %s\n",car.make);la $a0, string14addu $a1, $sp, 32addu $a1, $a1, 4jal printf

# 46 printf("model: %s\n",car.model);la $a0, string15addu $a1, $sp, 32addu $a1, $a1, 24jal printf

# 47 printf("year: %d\n",car.year);la $a0, string16lw $a1, 76($sp)jal printf

$ra<ra>

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]

make[0-3]

vehicleID

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

vehicleID

next

$a0vehicleID

$a1make[0-3]

$a2make[4-7]

$a3make[8-11] SP00

04

08

12

16

20

24

28

32

96

100

104

108

112116

120

124

<ra>


45

Why so much copying?

The program car.c passes the datastructure 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.


46

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



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);}


47

car-O0.s car2-O0.s

#include <stdio.h>#define STRINGLENGTH 20



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

main:subu $sp, 112sw $ra, 28($sp)



# 27 PrintCar(&mycar);addu $a0, $sp, 32jal PrintCar



48

car2-O0.s

main:subu $sp, 112sw $ra, 28($sp)



# 27 PrintCar(&mycar);addu $a0, $sp, 32jal PrintCar


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]

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

96

100

make[0-3]

vehicleID

$a0


49

car2-O0.sPrintCar:

subu $sp, 32sw $ra, 28($sp)sw $a0, 32($sp)

# 43 printf("vehicleID: %d\n",car->vehicleID);la $a0, $$13lw $t6, 32($sp)lw $a1, 0($t6)jal printf

# 44 printf("make: %s\n",car->make);la $a0, $$14lw $a1, 32($sp)addu $a1, $a1, 4jal printf

# 45 printf("model: %s\n",car->model);la $a0, $$15lw $a1, 32($sp)addu $a1, $a1, 24jal printf

# 46 printf("year: %d\n",car->year);la $a0, $$16lw $t7, 32($sp)lw $a1, 44($t7)jal printf

# 47 printf("mileage: %d\n",car->mileage);la $a0, $$17lw $t8, 32($sp)lw $a1, 48($t8)jal printf

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]

SP00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

96

100

make[0-3]

vehicleID

$a0


50

car2-O0.sPrintCar:


# 43 printf("vehicleID: %d\n",car->vehicleID);la $a0, string13lw $t6, 32($sp)lw $a1, 0($t6)jal printf

# 44 printf("make: %s\n",car->make);la $a0, string14lw $a1, 32($sp)addu $a1, $a1, 4jal printf

# 45 printf("model: %s\n",car->model);la $a0, string15lw $a1, 32($sp)addu $a1, $a1, 24jal printf

# 46 printf("year: %d\n",car->year);la $a0, string16lw $t7, 32($sp)lw $a1, 44($t7)jal printf

# 47 printf("mileage: %d\n",car->mileage);la $a0, string17lw $t8, 32($sp)lw $a1, 48($t8)jal printf

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]

SP

make[0-3]

vehicleID

$a0

$ra<ra>

00

04

08

12

16

20

24

28 <ra>

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

32

96

100

104

108

112

116

120124128

132

SP+64


51

car2-O0.sPrintCar:


# 43 printf("vehicleID: %d\n",car->vehicleID);la $a0, string13lw $t6, 32($sp)lw $a1, 0($t6)jal printf

# 44 printf("make: %s\n",car->make);la $a0, string14lw $a1, 32($sp)addu $a1, $a1, 4jal printf

# 45 printf("model: %s\n",car->model);la $a0, string15lw $a1, 32($sp)addu $a1, $a1, 24jal printf

# 46 printf("year: %d\n",car->year);la $a0, string16lw $t7, 32($sp)lw $a1, 44($t7)jal printf

# 47 printf("mileage: %d\n",car->mileage);la $a0, string17lw $t8, 32($sp)lw $a1, 48($t8)jal printf

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]

SP

make[0-3]

vehicleID

$a0

$ra<ra>

00

04

08

12

16

20

24

28 <ra>

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

32

96

100

104

108

112

116

120124128

132

SP+64

$t6SP+64

$a1vehicleID


52

Quiz #1

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

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

Where SP 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];


53

Quiz #1 - Solutionpredecessors and successors are two vectors of unsigned integers.predecessors’ first element is stored at SP+32successors’ first element is stored at SP+64

successors[5] = predecessors[7];

$t0 predecessors[7];successors[5] $t0;

$t1 Addr(predecessor[7]);$t0 0($t1);$t2 Addr(successors[5]);0($t2) $t0;

$t1 SP+32+7*4$t0 0($t1);$t2 SP+64+5*40($t2) $t0;

add $t1, $sp, 60lw $t0, 0($t1)add $t2, $sp, 84sw $t0,0($t2)

$t1 SP+60$t0 0($t1);$t2 SP+840($t2) $t0;


54

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 instructionsthat counts the number of characters in the string

whose pointer is at names[7].


55

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].

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

$t0 0;$t1 Address(names[7]);$t2 0($t1);while (*($t2) != 0) { $t0 = $t0+1; $t2 = $t2+1; }

$t0 0;$t2 names[7];while ( *($t2) != 0) { $t0 = $t0+1; $t2 = $t2+1; }

$t0 0;$t1 SP+16+7*4$t2 0($t1);while (0($t2) != 0) { $t0 = $t0+1; $t2 = $t2+1; }


56

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].

$t0 0;$t1 SP+16+7*4$t2 0($t1);while (0($t2) != 0) { $t0 = $t0+1; $t2 = $t2+1; }

add $t0, $zero, $zeroadd $t1, $sp, 44lw $t2, 0($t1)lb $t3,0($t2)beq $t3, $zero, doneaddi $t0, $t0, 1addi $t2, $t2, 1j loop;

loop:


57

Quiz #3

Consider the following C function:

Write the MIPS assembly code for the first statement of this function.Assume: temp $s0 node $s1

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

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


58

Quiz #3 - Solution

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

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

(*successors)[node] temp;

0($a1)[node] temp;

$t0 0($a1)$t1 Address($t0[node]) 0($t1) temp;

$t0 0($a1)$t0[node] temp;

$t0 0($a1)$t1 $t0 + 4*node0($t1) temp;

lw $t0, 0($a1)sll $t2, $s1, 2add $t1, $t0, $t2sw $s0,0($t1)


59

Quiz #4

Consider the following C function:

Write the MIPS assembly code to initialize the value of $a1 inthe 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]);}


60

How did we ended up with names ?

$a0A d a m \0

J a m e \0s

C e c i \0l y

M u r r \0a y


61

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

$a2

lw $t0, 0($a2) # $t0 names

$t0

A d a m \0

J a m e \0s

C e c i \0l y

M u r r \0a y


62


lw $t0, 0($a2) # $t0 names

$a2

$t3

lw $t0, 0($a2) # $t0 nameslw $t2, 8(sp) # $t2 nodesll $t2, $t2, 2 # $t2 4nodeadd $t3, $t0, $t2 # $t3 (names) + 4node

$t0

A d a m \0

J a m e \0s

C e c i \0l y

M u r r \0a y


63


A d a m \0

J a m e \0s

C e c i \0l y

$a2

lw $t0, 0($a2) # $t0 nameslw $t2, 8(sp) # $t2 nodesll $t2, $t2, 2 # $t2 4nodeadd $t3, $t0, $t2 # $t3 (names) + 4nodelw $a1, 0(t3) # $a1 (names)[node]

$t3$a1

$t0M u r r \0a y


64

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]);}

lw $t0, 0($a2) # $t0 nameslw $t2, 8(sp) # $t2 nodesll $t2, $t2, 2 # $t2 4nodeadd $t3, $t1, $t2 # $t3 (names) + 4nodelw $a1, 0(t3) # $a1 (names)[node]


65

Quiz #5

Write the code for the following C function:

/* NumOfStores counts the number of store instructions that appear between the instruction pointed to by FirstInstruction and the instruction pointed to by LastInstruction */

int NumberOfStores(unsigned int FirstInstruction, unsigned int LastInstruction){}

0x23 13 13 56OpCode rs rt address

sw $13, 56($13)

For this question, we will consider that the only type of store operationavailable in the processor is a store word, such as:

100011 01101 01101 0000 0000 0011 100031 26 15 020 1625 21


66

Quiz #5

/* NumOfStores counts the number of store instructions that appear between the instruction pointed to by FirstInstruction and the instruction pointed to by LastInstruction */

int NumberOfStores(unsigned int FirstInstruction, unsigned int LastInstruction){ unsigned int *instruction; int count = 0; if(LastInstruction < FirstInstruction) return 0; for(instruction = FirstInstruction ; instruction <= LastInstruction ; instruction++) if((*instruction & 0xfc000000) == 0x8c000000) count++; return count;}

0x23 13 13 56OpCode rs rt address

sw $13, 56($13)

100011 01101 01101 0000 0000 0011 100031 26 15 020 1625 21


67

Quiz #6

For this quiz, assume that the only instructions that can effectcontrol flow program transfers are:

bne $s3, $s4, 8 PC PC + 4 if($1 $2) PC PC + 8

000101 10011 10100 0000 0000 0000 100031 26 15 020 1625 21

5 19 20 8OpCode rs rt address

j Exit PC concat(PC[31-28],IR[25-0])<<2

000010 00 0000 0000 0000 0000 0000 010031 26 25 0

2 4OpCode address


68

Quiz #6Assume that the following C functions are available for your use in thisquiz:

/* The following functions can manipulate a bit vector of arbitrary length */unsigned int *CreateBitvector(unsigned int VectorLenght);void SetBit(unsigned int *BitVector, unsigned int BitNumber);void ResetBitRange(unsigned int *BitVector, unsigned int FirstBit, unsigned int LastBit);void ResetBit(unsigned int *BitVector, unsigned int BitNumber);

/* BBLeaders returns a bit vector where the bit corresponding to instruction i is 1 if and only if i is a basic block leader. */

unsigned int *BBLeaders(unsigned int *FirstInstruction, unsigned int *LastInstruction){}

Write the BBLeaders function:


69

Quiz #6

/* BBLeaders returns a bit vector where the bit corresponding to instruction i is 1 if and only if i is a basic block leader. */void *BBLeaders(void *FirstInstruction, void *LastInstruction){ unsigned int *instruction; unsigned int *bitvector; int bitcount = 0; unsigned int opcode; unsigned int num_instructions;

if(LastInstruction <= FirstInstruction) return 0; num_instructions = LastInstruction - FirstInstruction +1; bitvector = CreateBitVector(num_instructions); ResetBitRange(bitvector,0,num_instructions-1); SetBit(bitvector,0); /* first instruction is a leader */ for(instruction = (unsigned int *) FirstInstruction ; instruction <= LastInstruction ; instruction++) { opcode = *instruction & 0xfc000000 if(opcode == 0x14000000)

{ /* process the branch instruction */ }

if(opcode == 0x08000000){ /* process jump */}

bitcount++; } return ((void *) bitvector);}


70

Quiz #7Write MIPS assembly for thefollowing program.Assume that:

apple $sp+8 ptr $sp+4ind $sp+0

# 7 ind = &ptr;addi $t0, $sp, 4 # $t0 &ptr sw $t0, 0($sp) # ind &ptr# 8 *ind = &apple;lw $t0, 0($sp)addi $t1, $sp,8 # $t1 &applesw $t1, 0($t0) # *ind &apple# 9 **ind = 123;addi $t3,$zero,123lw $t4, 0($sp) # $t4 *indlw $t5, 0($t4) # $t5 **indsw $t3, 0($t5) # **ind 123# 10 ind++;lw $t0, 0($sp)addi $t0, $t0, 4sw $t0, 0($sp) # ind++# 11 *ptr++;lw $t6, 4($sp) # $t6 ptrlw $t7, 0($t6) # $t7 *ptraddi $t7, $t7, 1sw $t7, 0($t6) # *ptr++# 12 apple++;lw $t8, 8(sp) # $t8 appleaddi $t8, $t8, 1sw $t8, 8(sp) # apple++

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

Stackappleptr

$sp 0

4

8


71

Quiz #7# 7 ind = &ptr;addi $t0, $sp, 4 # $t0 &ptr sw $t0, 0($sp) # ind &ptr# 8 *ind = &apple;lw $t0, 0($sp)addi $t1, $sp,8 # $t1 &applesw $t1, 0($t0) # *ind &apple# 9 **ind = 123;addi $t3,$zero,123lw $t4, 0($sp) # $t4 *indlw $t5, 0($t4) # $t5 **indsw $t3, 0($t5) # **ind 123# 10 ind++;lw $t0, 0($sp)addi $t0, $t0, 4sw $t0, 0($sp) # ind++# 11 *ptr++;lw $t6, 4($sp) # $t6 ptrlw $t7, 0($t6) # $t7 *ptraddi $t7, $t7, 1sw $t7, 0($t6) # *ptr++# 12 apple++;lw $t8, 8(sp) # $t8 appleaddi $t8, $t8, 1sw $t8, 8(sp) # apple++




$sp+4

Stackappleptr

$sp 0

4

8

What are the values in the stackafter line 12 is executed?Assume that:



72





$sp+4

Stackappleptr

$sp 0

4

8




73





$sp+4

Stackapple$sp+8

$sp 0

4

8




74





$sp+4

Stackapple$sp+8

$sp 0

4

8




75





$sp+4

Stack123

$sp+8$sp 0

4

8




76





$sp+4

Stack123

$sp+8$sp 0

4

8




77





$sp+8

Stack123

$sp+8$sp 0

4

8




78





$sp+8

Stack123

$sp+8$sp 0

4

8




79





$sp+8

Stack124

$sp+8$sp 0

4

8




80





$sp+8

Stack124

$sp+8$sp 0

4

8




81





$sp+8

Stack125

$sp+8$sp 0

4

8




82





$sp+8

Stack125

$sp+8$sp 0

4

8



cmput229 - fall 2003

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