Machine-Level%20Programming%20IV:%20Structured%20Data%20Apr.%2023,%202008

1
Machine-Level Programming IVStructured
DataApr. 23, 2008
EECS213

• Topics
• Arrays
• Structures
• Unions

2
Basic Data Types

• Integral
• Stored operated on in general registers
• Intel GAS Bytes C
• byte b 1 unsigned char
• word w 2 unsigned short
• double word l 4 unsigned int
• Floating Point
• Stored operated on in floating point registers
• Intel GAS Bytes C
• Single s 4 float
• Double l 8 double
• Extended t 10/12 long double

3
Array Allocation

• Basic Principle
• T AL
• Array of data type T and length L
• Contiguously allocated region of L sizeof(T)
  bytes

char p3
4
Array Access

• Basic Principle
• T AL
• Array of data type T and length L
• Identifier A can be used as a pointer to array
  element 0
• Reference Type Value
• val4 int 3
• val int x
• val1 int x 4
• val2 int x 8
• val5 int ??
• (val1) int 5
• val i int x 4 i

5
Array Example
typedef int zip_dig5 zip_dig cmu 1, 5, 2,
1, 3 zip_dig mit 0, 2, 1, 3, 9 zip_dig
ucb 9, 4, 7, 2, 0

• Notes
• Declaration zip_dig cmu equivalent to int
  cmu5
• Example arrays were allocated in successive 20
  byte blocks
• Not guaranteed to happen in general

6
Array Accessing Example

• Computation
• Register edx contains starting address of array
• Register eax contains array index
• Desired digit at 4eax edx
• Use memory reference (edx,eax,4)

int get_digit (zip_dig z, int dig) return
zdig

• Memory Reference Code

edx z eax dig movl
(edx,eax,4),eax zdig
7
Referencing Examples

• Code Does Not Do Any Bounds Checking!
• Reference Address Value Guaranteed?
• mit3 36 4 3 48 3
• mit5 36 4 5 56 9
• mit-1 36 4-1 32 3
• cmu15 16 415 76 ??
• Out of range behavior implementation-dependent
• No guaranteed relative allocation of different
  arrays

Yes
No
No
No
8
Array Loop Example
int zd2int(zip_dig z) int i int zi 0
for (i 0 i lt 5 i) zi 10 zi
zi return zi

• Original Source

• Transformed Version
• As generated by GCC
• Eliminate loop variable i
• Convert array code to pointer code
• Express in do-while form
• No need to test at entrance

int zd2int(zip_dig z) int zi 0 int zend
z 4 do zi 10 zi z z
while(z lt zend) return zi
9
Array Loop Implementation

• Registers
• ecx z
• eax zi
• ebx zend
• Computations
• 10zi z implemented as z 2(zi4zi)
• z increments by 4

int zd2int(zip_dig z) int zi 0 int zend
z 4 do zi 10 zi z z
while(z lt zend) return zi
int zd2int(zip_dig z) int zi 0 int zend
z 4 do zi 10 zi z z
while(z lt zend) return zi
int zd2int(zip_dig z) int zi 0 int zend
z 4 do zi 10 zi z z
while(z lt zend) return zi
int zd2int(zip_dig z) int zi 0 int zend
z 4 do zi 10 zi z z
while(z lt zend) return zi
int zd2int(zip_dig z) int zi 0 int zend
z 4 do zi 10 zi z z
while(z lt zend) return zi
ecx z xorl eax,eax zi 0 leal
16(ecx),ebx zend z4 .L59 leal
(eax,eax,4),edx 5zi movl (ecx),eax
z addl 4,ecx z leal (eax,edx,2),eax
zi z 2(5zi) cmpl ebx,ecx z
zend jle .L59 if lt goto loop
ecx z xorl eax,eax zi 0 leal
16(ecx),ebx zend z4 .L59 leal
(eax,eax,4),edx 5zi movl (ecx),eax
z addl 4,ecx z leal (eax,edx,2),eax
zi z 2(5zi) cmpl ebx,ecx z
zend jle .L59 if lt goto loop
ecx z xorl eax,eax zi 0 leal
16(ecx),ebx zend z4 .L59 leal
(eax,eax,4),edx 5zi movl (ecx),eax
z addl 4,ecx z leal (eax,edx,2),eax
zi z 2(5zi) cmpl ebx,ecx z
zend jle .L59 if lt goto loop
ecx z xorl eax,eax zi 0 leal
16(ecx),ebx zend z4 .L59 leal
(eax,eax,4),edx 5zi movl (ecx),eax
z addl 4,ecx z leal (eax,edx,2),eax
zi z 2(5zi) cmpl ebx,ecx z
zend jle .L59 if lt goto loop
ecx z xorl eax,eax zi 0 leal
16(ecx),ebx zend z4 .L59 leal
(eax,eax,4),edx 5zi movl (ecx),eax
z addl 4,ecx z leal (eax,edx,2),eax
zi z 2(5zi) cmpl ebx,ecx z
zend jle .L59 if lt goto loop
10
Array reference quiz

• Assume short S, int i
• edx S0, ecx i
• For each expression, give its type and the
  assembly code to store its value in eax, if
  pointer, or ax if short

Expression Type Code
S1
S3
Si
S4i1
Si-5
11
Nested Array Example
define PCOUNT 4 zip_dig pghPCOUNT 1, 5,
2, 0, 6, 1, 5, 2, 1, 3 , 1, 5, 2, 1, 7
, 1, 5, 2, 2, 1

• Declaration zip_dig pgh4 equivalent to int
  pgh45
• Variable pgh denotes array of 4 elements
• Allocated contiguously
• Each element is an array of 5 ints
• Allocated contiguously
• Row-Major ordering of all elements guaranteed

12
Nested Array Allocation

• Declaration
• T ARC
• Array of data type T
• R rows, C columns
• Type T element requires K bytes
• Array Size
• R C K bytes
• Arrangement
• Row-Major Ordering

int ARC
4RC Bytes
13
Nested Array Row Access

• Row Vectors
• Ai is array of C elements
• Each element of type T
• Starting address A i C K

int ARC
  
  
A
AiC4
A(R-1)C4
14
Nested Array Row Access Code
int get_pgh_zip(int index) return
pghindex

• Row Vector
• pghindex is array of 5 ints
• Starting address pgh20index
• Code
• Computes and returns address
• Compute as pgh 4(index4index)

eax index leal (eax,eax,4),eax 5
index leal pgh(,eax,4),eax pgh (20 index)
15
Nested Array Element Access

• Array Elements
• Aij is element of type T
• Address A (i C j) K

A i j
int ARC
Ai
  
  
A i j
  
  
A
AiC4
A(R-1)C4
A(iCj)4
16
Nested Array Element Access Code

• Array Elements
• pghindexdig is int
• Address
• pgh 20index 4dig
• Code
• Computes address
• pgh 4dig 4(index4index)
• movl performs memory reference

int get_pgh_digit (int index, int dig)
return pghindexdig
ecx dig eax index leal
0(,ecx,4),edx 4dig leal (eax,eax,4),eax
5index movl pgh(edx,eax,4),eax (pgh
4dig 20index)
17
Strange Referencing Examples
zip_dig pgh4

• Reference Address Value Guaranteed?
• pgh33 7620343 148 2
• pgh25 7620245 136 1
• pgh2-1 762024-1 112 3
• pgh4-1 762044-1 152 1
• pgh019 76200419 152 1
• pgh0-1 762004-1 72 ??
• Code does not do any bounds checking
• Ordering of elements within array guaranteed

Yes
Yes
Yes
Yes
Yes
No
18
Multi-Level Array Example

• Variable univ denotes array of 3 elements
• Each element is a pointer
• 4 bytes
• Each pointer points to array of ints

zip_dig cmu 1, 5, 2, 1, 3 zip_dig mit
0, 2, 1, 3, 9 zip_dig ucb 9, 4, 7, 2, 0
define UCOUNT 3 int univUCOUNT mit, cmu,
ucb
19
Element Access in Multi-Level Array
int get_univ_digit (int index, int dig)
return univindexdig

• Computation
• Element access MemMemuniv4index4dig
• Must do two memory reads
• First get pointer to row array
• Then access element within array

ecx index eax dig leal
0(,ecx,4),edx 4index movl univ(edx),edx
Memuniv4index movl (edx,eax,4),eax
Mem...4dig
20
Array Element Accesses

• Similar C references
• Nested Array
• Element at
• Mempgh20index4dig

• Different address computation
• Multi-Level Array
• Element at
• MemMemuniv4index4dig

int get_pgh_digit (int index, int dig)
return pghindexdig
int get_univ_digit (int index, int dig)
return univindexdig
21
Strange Referencing Examples

• Reference Address Value Guaranteed?
• univ23 5643 68 2
• univ15 1645 36 0
• univ2-1 564-1 52 9
• univ3-1 ?? ??
• univ112 16412 64 7
• Code does not do any bounds checking
• Ordering of elements in different arrays not
  guaranteed

Yes
No
No
No
No
22
Using Nested Arrays
define N 16 typedef int fix_matrixNN

• Strengths
• C compiler handles doubly subscripted arrays
• Generates very efficient code
• Avoids multiply in index computation
• Limitation
• Only works if have fixed array size

/ Compute element i,k of fixed matrix product
/ int fix_prod_ele (fix_matrix a, fix_matrix b,
int i, int k) int j int result 0 for
(j 0 j lt N j) result
aijbjk return result
23
Dynamic Nested Arrays

• Strength
• Can create matrix of arbitrary size
• Programming
• Must do index computation explicitly
• Performance
• Accessing single element costly
• Must do multiplication

int new_var_matrix(int n) return (int )
calloc(sizeof(int), nn)
int var_ele (int a, int i, int j, int n)
return ainj
movl 12(ebp),eax i movl 8(ebp),edx
a imull 20(ebp),eax ni addl
16(ebp),eax nij movl (edx,eax,4),eax
Mema4(inj)
24
Dynamic Array Multiplication

• Without Optimizations
• Multiplies
• 2 for subscripts
• 1 for data
• Adds
• 2 for array indexing
• 1 for loop index
• 1 for data

/ Compute element i,k of variable matrix
product / int var_prod_ele (int a, int b,
int i, int k, int n) int j int result
0 for (j 0 j lt n j) result
ainj bjnk return result
25
Optimizing Dynamic Array Mult.
int j int result 0 for (j 0 j lt n
j) result ainj bjnk
return result

• Optimizations
• Performed when set optimization level to -O2
• Code Motion
• Expression in can be computed outside loop
• Strength Reduction
• Incrementing j has effect of incrementing jnk
  by n
• Performance
• Compiler can optimize regular access patterns

int j int result 0 int iTn in
int jTnPk k for (j 0 j lt n j)
result aiTnj bjTnPk jTnPk
n return result
26
Structures

• Concept
• Contiguously-allocated region of memory
• Refer to members within structure by names
• Members may be of different types
• Accessing Structure Member

struct rec int i int a3 int p
Memory Layout
Assembly
void set_i(struct rec r, int val)
r-gti val
eax val edx r movl eax,(edx)
Memr val
27
Generating Pointer to Struct. Member
r
struct rec int i int a3 int p
i
a
p
0
4
16
r 4 4idx

• Generating Pointer to Array Element
• Offset of each structure member determined at
  compile time

int find_a (struct rec r, int idx) return
r-gtaidx
ecx idx edx r leal 0(,ecx,4),eax
4idx leal 4(eax,edx),eax r4idx4
28
Structure Referencing (Cont.)

• C Code

struct rec int i int a3 int p
void set_p(struct rec r) r-gtp
r-gtar-gti
edx r movl (edx),ecx r-gti leal
0(,ecx,4),eax 4(r-gti) leal
4(edx,eax),eax r44(r-gti) movl
eax,16(edx) Update r-gtp
29
Structure reference quiz

• Offsets in bytes for
• p
• s.x
• s.y
• next
• Total bytes for prob?

• struct prob
• int p
• struct
• int x
• int y
• s
• struct prob next

30
Structure reference quiz

• Offsets in bytes for
• p
• s.x
• s.y
• next
• Total bytes for prob?

• struct prob
• int p
• struct
• int x
• int y
• s
• struct prob next

31
Structure reference quiz

• Offsets in bytes for
• p 0
• s.x 4
• s.y 8
• next 12
• Total bytes for prob? 16

• struct prob
• int p
• struct
• int x
• int y
• s
• struct prob next

32
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax ____________1______________ m
ovl 8(eax), edx ____________2______________ mo
vl edx, 4(eax) ____________3______________ lea
l 4(eax), edx ____________4______________ movl
edx, (eax) ____________5______________ movl
eax, 12(eax) ____________6______________
33
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx ____________2______________ movl edx,
4(eax) ____________3______________ leal
4(eax), edx ____________4______________ movl
edx, (eax) ____________5______________ movl
eax, 12(eax) ____________6______________
34
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
____________3______________ leal 4(eax), edx
____________4______________ movl edx, (eax)
____________5______________ movl eax, 12(eax)
____________6______________
35
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx
____________4______________ movl edx, (eax)
____________5______________ movl eax, 12(eax)
____________6______________
36
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx put
(sp-gts.x) in edx movl edx, (eax)
____________5______________ movl eax, 12(eax)
____________6______________
37
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx put
(sp-gts.x) in edx movl edx, (eax) put
(sp-gts.x) in sp-gtp movl eax, 12(eax)
____________6______________
38
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
____7_____ sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx put
(sp-gts.y) in edx movl edx, (eax) put
(sp-gts.y) in sp-gtp movl eax, 12(eax) put sp
in sp-gtnext
39
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
sp-gts.y sp-gtp ____8_____ sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx put
(sp-gts.y) in edx movl edx, (eax) put
(sp-gts.y) in sp-gtp movl eax, 12(eax) put sp
in sp-gtnext
40
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
sp-gts.y sp-gtp (sp-gts.x) sp-gtnext
____9_____
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx put
(sp-gts.y) in edx movl edx, (eax) put
(sp-gts.y) in sp-gtp movl eax, 12(eax) put sp
in sp-gtnext
41
Structure reference quiz
void sp_init (struct prob sp) sp-gts.x
sp-gts.y sp-gtp (sp-gts.x) sp-gtnext
sp
movl 8(ebp), eax put sp in eax movl 8(eax),
edx put sp-gts.y in edx movl edx, 4(eax)
put sp-gts.y in sp-gts.x leal 4(eax), edx put
(sp-gts.y) in edx movl edx, (eax) put
(sp-gts.y) in sp-gtp movl eax, 12(eax) put sp
in sp-gtnext