CS 208 s21 — Data Structure Representation: structs

Table of Contents

1 Video

Here is a video lecture for the material outlined below. It covers CSPP section 3.9.1 and 3.9.3 (p. 265–267, 273–275). It contains sections on

  1. multi-level arrays (0:27)
  2. spreadsheet visualization (2:41)
  3. assembly for multi-level array access (5:12)
  4. struct basics (13:40)
  5. accessing struct fields (18:59)
  6. size of a struct (22:05)
  7. assembly for struct access (24:22)
  8. data alignment (29:41)
  9. alignment for structs (36:28)

The Panopto viewer has table of contents entries for these sections. Link to the Panopto viewer: https://carleton.hosted.panopto.com/Panopto/Pages/Viewer.aspx?id=d44ba1b1-cdb0-4cb4-b6e6-ac500113c492

2 Multilevel Arrays

  • is this multi-dimensional array equivalent to previous sea?
int sea0[5] = {9, 8, 1, 9, 5};
int sea1[5] = {9, 8, 1, 0, 5};
int sea2[5] = {9, 8, 1, 0, 3};
int sea3[5] = {9, 8, 1, 1, 5};
int *sea_m[4] = {sea0, sea1, sea2, sea3};
  • it contains the same 20 ints
  • however, each of the four elements of sea_m is a pointer—none of the elements of sea were pointers
  • within each of the rows (sea0, sea1, sea2, sea3), the 5 ints are allocated as a contiguous block of memory, but each row could be put anywhere
  • see the difference visually in this spreadsheet
  • the C code for get_sea_m_digit is the same as get_sea_digit for 2D arrays
int get_sea_m_digit (int index, int digit) {
    return sea_m[index][digit];
}
  • but the assembly for accessing an element will be different
get_sea_digit:
        leaq (%rdi,%rdi,4), %rax  # 5 * index
        addl %rax, %rsi           # 5 * index + digit
        movl sea(,%rsi,4), %eax   # *(sea + 4 * (5 * index + digit))
        ret

get_sea_m_digit:
        salq $2, %rsi # rsi = 4*digit
        addq sea_m(,%rdi,8), %rsi # p = sea_m[index] + 4*digit
        movl (%rsi), %eax # return *p
        ret
  • accessing an element now requies two memory accesses
  • the benefit of this multilevel structure is that the rows can be different lengths
  • array access looks the same sea[3][2] and sea_m[3][2], but underneath
    • Mem[ sea + 20*index + 4*digit ] vs Mem[ Mem[ sea_m + 8*index ] + 4*digit ]

3 Structures

Two ways to create data types in C: structures (struct) and unions (union) (we won't worry about unions in this course)

// a way of combining different types of data together
struct song {
    char *title;
    int length_in_seconds;
    int year_released;
};
struct song song1;
song1.title = "What is Urinetown?";
song1.length_in_seconds = 213;
song1.year_released = 2001;
  • variable declarations like any other type:
    • struct name name1;
    • struct name *pn;
    • struct name name_ar[3];
  • common to use typedef to give the struct type a more concise alias
    • typedef struct song song_t; makes it so we can use song_t anywhere we would need to use struct song
  • access fields with . or -> in the case of the pointer (p->field is shorthand for (*p).field)
  • like arrays, struct elements are stored in a contiguous region with a pointer to the first byte
    • what will sizeof(struct song) return? (sizeof is a built-in operator that returns the size, in bytes, of a type)
      • 16 bytes
    • compiler maintains the byte offset information needed to access additional fields (e.g., length_in_seconds is 8 bytes from the start of the struct)
    • can find offset of individual fields using offsetof(type, member)

3.1 Examples

struct rec {
    int a[4];
    long i;
    struct rec *next;
};

struct-rec.png

  • fields are laid out in memory in same order in which they were declared
  • machine code knows nothing about structures, all just byte offsets from a pointer
long get_i(struct rec *r) {
    return r->i;
}
  • the compiler knows the field i is always 16 bytes (the size of the array of 4 ints a) from the start of the struct:
get_i:
        movq 16(%rdi), %rax
        ret
long* addr_of_i(struct rec *r) {
    return &(r->i);
}
  • note to get the address of the i field, we use lea to add 16 to the address of the struct
addr_of_i:
        leaq 16(%rdi), %rax
        ret
struct rec** addr_of_next(struct rec *r) {
    return &(r->next);
}
  • the next field is 24 bytes from the start of the struct (16 for a + 8 for i)
addr_of_next:
        leaq 24(%rdi), %rax
        ret
int get_array_elem (struct rec *r, long index) {
    return r->a[index];
}
  • since a is located at the start of the struct, accessing an element requires no offset, just normal array indexing
get_array_elem:
        movl (%rdi, %rsi, 4), %eax
        ret

3.2 Data Alignment

  • suppose a processor always fetches 8 bytes from an address that must be a multiple of 8
    • if every double is guaranteed to have a memory address that is a multiple of 8, then it's guaranteed to take only a single operation to read
    • otherwise, it would take two operations if a double were split across two 8-byte blocks
  • this kind of behavior is typical of hardware interfacing between the processor and memory
    • hence, systems institute alignment restrictions to improve memory performance
    • Intel recommends data alignment to improve performance
  • x86-64 alignment principle: any primitive object of \(K\) bytes must have an address that is a multiple of \(K\)
  • this means for structures, the compiler sometimes must insert gaps between fields to maintain alignment (internal fragmentation)
    • even if this padding isn't required within a structure, it sometimes must be added to the end to ensure an array of structures is aligned (external fragmentation)
      • each structure has alignment requirement \(K_{max}\) = largest alignment of any element
      • counts array elements individually as elements