Machine Programming Data

CSC 235 - Computer Organization

References

Slides adapted from CMU

Outline

Arrays
- One-dimensional
- Two-dimensional
- Multi-level
Structures
- Allocation
- Access
- Alignment

Array Allocation

Basic Principle
- C syntax: T A[L]
- Array of type T and length L
- Contiguously allocated region of L * sizeof(T) bytes in memory
Examples:
- char string[12]: 12 * sizeof(char) = 12 bytes
- int val[5]: 5 * sizeof(int) = 20 bytes

Array Access

Basic Principle
- The array identifier can be used as a pointer to array element 0: type T*
Examples: int val[5] = {1, 5, 2, 3, 1};

Reference Type Value

val[4] int 1

val int * x

val+1 int * x+4

&val[2] int * x+8

*(val+1) int 5

where x is the address of the first byte of val

Reference	Type	Value
`val[4]`	`int`	`1`
`val`	`int *`	`x`
`val+1`	`int *`	`x+4`
`&val[2]`	`int *`	`x+8`
`*(val+1)`	`int`	`5`

Array Example

C code

#define ZLEN 5
typedef int zip_dig[ZLEN];

zip_dig_ku = {1, 9, 5, 3, 0};

int get_digit (zip_dig z, int digit) {
   return z[digit];
}

Array Example

Assembly code

  # %rdi = z (starting address)
  # %rsi = digit (index)
movl (%rdi, %rsi, 4), %eax  # z[digit]

Array Loop Example

C code

void zincr(zip_dig z) {
    size_t i;
    for (i = 0; i < ZLEN; i++) {
        z[i]++;
    }
}

Array Loop Example

Assembly code

    # %rdi = z
    movl    $0, %eax             #   i = 0
    jmp     .L3                  #   goto middle
.L4:                             # loop:
    addl    $1, (%rdi, %rax, 4)  #   z[i]++
    addq    $1, %rax             #   i++
.L3:                             # middle:
    cmpq    $4, %rax             #   i:4
    jbe     .L4                  #   if <=, goto loop
    rep; ret

Multidimensional (Nested) Arrays

Declaration: T A[R][C];
- 2D array of type T
- R rows, C columns
Array Size
- R * C * sizeof(T)
Arrangement
- Row-major ordering

\[ \begin{bmatrix} \texttt{A[0][0]} & \dots & \texttt{A[0][C-1]}\\ \vdots & \ddots & \vdots\\ \texttt{A[R-1][0]} & \dots & \texttt{A[R-1][C-1]}\\ \end{bmatrix} \]

Nested Array Example

C code

#define PCOUNT 4
typedef int zip_dig[5];

zip_dig pgh[PCOUNT] =
  {{1, 9, 5, 1, 1},
   {1, 9, 5, 3, 6},
   {1, 9, 5, 6, 2},
   {1, 9, 5, 2, 9}};

Nested Array Row Access

Row Vectors
- A[i] is array C elements of type T
- Starting address A + i * (C * sizeof(T))
Array Elements
- A[i][j] is element of type T, which requies K bytes
- Address A + i * (C* K) + j * k \(\rightarrow\) A + (i * C + j) * K

Nested Array Row Access Code

C code
```
int *get_pgh_zip(int index) {
    return pgh[index];
}
```
- pgh[index] is array of 5 int values with starting address pgh + 20 * index

Assembly code

# %rdi = index
leaq (%rdi, %rdi, 4), %rax  # 5 + index
leaq pgh(,%rax,4), %rax     # pgh + (20 * index)

Computes return address as pgh + 4 * (index + 4 * index)

Nested Array Element Access Code

C code

int get_pgh_digit(int index, int dig) {
    return pgh[index][dig];
}

Assembly code

leaq (%rdi, %rdi, 4), %rax  # 5 + index
addl %rax, %rsi             # 5 * index + dig
leaq pgh(,%rsi,4), %rax     # M[pgh + 4 * (5 * index + dig)]

Array elements
- pgh[index][digit] is a value of type int
- Address: pgh + 20 * index + 4 * dig

Multi-Level Array Example

C code

zip_dig b = { 1, 9, 5, 1, 1};
zip_dig k = { 1, 9, 5, 3, 0};
zip_dig t = { 1, 9, 5, 6, 2};

#define TCOUNT 3
int *town[TCOUNT] = {b, k, t}

Variable town denotes an array of 3 elements
Each element is a pointer (8 bytes)
Each pointer points to array of int values

Element Access in Multi-Level Array

C Code

int get_town_digit(size_t index, size_t digit) {
  return town[index][digit]
}

Assembly Code

salq $2, %rsi               # 4 * digit
addq town(, %rdi, 8), %rsi  # p = town[index] + 4 * digit
movl (%rsi), %eax           # return *p

Computation
- Element access Mem[Mem[town+8*index]+4*digit]
- Must perform two memory reads

N \(\times\) N Matrix Code

Fixed dimensions (know at compile time)

#define N 16
typedef int fix_matrix[N][N];

/* Get element at A[i][j] */
int fix_ele(fix_matrix A, size_t i, size_t j) {
    return A[i][j];
}

N \(\times\) N Matrix Code

Variable dimensions, explicit indexing

#define IDX(n, i, j) ((i)*(n)+(j))

/* Get element at A[i][j] */
int vec_ele(size_t n, int *A, size_t i, size_t j) {
    return A[IDX(n,i,j)];
}

N \(\times\) N Matrix Code

Variable dimensions, implicit indexing (now supported by gcc)

/* Get element at A[i][j] */
int vec_ele(size_t n, int A[n][n], size_t i, size_t j) {
    return A[i][j];
}

16 \(\times\) 16 Fixed Matrix Access

Array elements
- int A[16][16];
- Address A + i * (C * K) + j * K
- C = 16, K = 4

Assembly Code

# A in %rdi, i in %rsi, j in %rdx
salq  $6, %rsi               # 64 * i
addq  %rsi, %rdi             # A + 64 * i
movl  (%rdi, %rdx, 4), %eax  # Mem[A + 64*i + 4*j]
ret

n \(\times\) n Variable Matrix Access

Array elements
- size_t n;
- int A[n][n];
- Address A + i * (C * K) + j * K
- C = n, K = 4
- Must perform integer multiplication

Assembly Code

# n in %rdi, A in %rsi, i in %rdx, j in %rcx
imulq  %rdx, %rdi             # n * i
leaq   (%rsi, %rdi, 4), %rax  # A + 4*n*i
movl   (%rax, %rcx, 4), %eax  # A + 4*n*i + 4*j
ret

Structure Representation

Structure represented as block of memory big enough to hold all the fields
Fields ordered according to declaration
- Even if another ordering could yield a more compact representation
Compiler determines overall size + positions of fields

Example

struct rec {
  int a[4];         // 16 bytes
  size_t i;         // 8 bytes
  struct rec *next; // 8 bytes
}

Generating Pointer to Structure Member

C code

int *get_ap(struct rec *r, size_t, idx) {
  return &r->a[idx];
}

Assembly code
```
# r in %rdi, idx in %rsi
leaq (%rdi, %rsi, 4), %rax
ret
```
- Offset of each structure member is determined at compile time
- Compute as r + 4 * idx

Following Linked List Example

C Code

long length(struct rec* r) {
    long len = 0L;
    while (r) {
        len++;
        r = r->next;
    }
    return len;
}

Loop Assembly Code

.L11:                  # loop:
  addq  $1, %rax       #   len++
  movq  24(%rdi), %rdi #   r = Mem[r+24]
  testq %rdi, %rdi     #   test r
  jne   .L11           #   if != 0 goto loop

Following Linked List Example 2

C Code

void set_val(struct rec* r, int val) {
    while (r) {
        size_t i = r->i;
        // no bounds check
        r->a[i] = val;
        r = r->next;
    }
}

Loop Assembly Code

.L11:                         # loop:
  movq  16(%rdi), %rax        #   i = Mem[r+16]
  movl  %esi, (%rdi, %rax, 4) #   Mem[r+4*i] = val
  movq  24(%rdi), %rdi        #   r = Mem[r+24]
  testq %rdi, %rdi            #   test r
  jne   .L11                  #   if != 0 goto loop

Structures and Alignment

Example Code

struct S1 {
    char c;   // 1 byte
    int i[2]; // 2*4 bytes
    double v; // 8 bytes
} *p;

Unaligned Data
- c starts at address p
- i[0] starts at address p+1
- i[1] starts at address p+5
- v starts at address p+9

Structures and Alignment

Example Code

struct S1 {
    char c;   // 1 byte
    int i[2]; // 2*4 bytes
    double v; // 8 bytes
} *p;

Aligned Data
- Primitive data type requires B bytes implies address must be mutliple of B
- c starts at address p+0
- i[0] starts at address p+4 (3-byte gap)
- i[1] starts at address p+8
- v starts at address p+16 (4-byte gap)

Alignment Principles

Aligned Data
- Primitive data type requires B bytes implies address must be mutliple of B
- Required on some machines; advised on x86-64
Motivation for Aligning Data
- Memory is accessed by (aligned) chunks of 4 or 8 bytes (system dependent)
- Inefficient to load or store data that spans chace lines
- Virtual memory is tricky when data spans 2 pages
Compiler inserts gaps in structure to ensure correct alignmnet of fields.

Specific Cases of Alignment (x86-64)

1 byte: char, \(\ldots\)
- no restrictions on address
2 bytes: short, \(\ldots\)
- lowest 1 bit of address must be 0
4 bytes: int, float, \(\ldots\)
- lowest 2 bits address must be 00
8 bytes: double, long, char *, \(\ldots\)
- lowest 3 bits address must be 000

Satisfying Alignment with Structures

Within structure:
- must satisfy each element’s alignment requirement
Overall structure placement
- Each structure has alignment requirement K
  - K = largest alignment of any element
- Initial address and structure length must be multiples of K

Example:

K = 8 due to double element

struct S1 {
    char c;   // 1 byte
    int i[2]; // 2*4 bytes
    double v; // 8 bytes
} *p;

Meeting Overall Alignment Requirement

For largest alignment requirement K
Overall structure must be multiple of K

Example

struct S2 {
    double v; // 8 bytes
    int i[2]; // 2*4 bytes
    char c;   // 1 byte
} *p;

Since the char is last, the overall structure needs 7 bytes of padding at the end.

Arrays of Structures

Overall structure length multiple of K
Satisfy alignment requirement for every element.

Example:

struct S2 {
    double v; // 8 bytes
    int i[2]; // 2*4 bytes
    char c;   // 1 byte
} a[10];

Each array element is aligned to 24 bytes.

Accessing Array Elements

C code

struct S3 {
    short i; // 2 bytes (2 byte gap)
    float v; // 4 bytes
    short j; // 2 bytes (2 byte padding)
} a[10];

short get_j(int idx) {
    return a[idx].j;
}

Assembly code

# %rdi = idx
leaq (%rdi, %rdi, 2), %rax  # 3*idx
movzwl a+8(,%rax,4), %eax   # a+8 resolved at link time

Saving Space in Structures

Put large data types first

Example 1 (12 bytes total):

struct S4 {
  char c; // 1 byte (3 byte gap)
  int i;  // 4 bytes
  char d; // 1 byte (3 bytes padding)
} *p;

Example 2 (8 bytes total):

struct S5 {
  int i;  // 4 bytes
  char c; // 1 byte (no gap)
  char d; // 1 byte (2 bytes padding)
} *p;

Summary

Arrays
- One-dimensional
- Two-dimensional
- Multi-level
Structures
- Allocation
- Access
- Alignment