Heterogeneous data structure

Structs

• Consider the following structure declaration:

  struct rec {
      int i;
      int j;
      int a[2];
      int *p;
  }
  
  struct rec *r;

  

  • Suppose variable r is in register %rdi, the following code copies element r->i to element r->j

    movl (%rdi), %eax // get r->i
    movl %eax, 4(%rdi) // Store in r->j

  • The offset of field i is 0, the address of this field is simply the value of r.

  • The offset of field j is 4, the address of this field = address of r + offset 4.

  • Suppose variable r is in register %rdi, long integer variable i is in register %rsi, we can generate the pointer value &(r->a[i]) with the single instruction:

    leaq 8(%rdi,%rsi,4), %rax //Set %rax to &r->a[i]

• Example 2:

  struct test {
    short *p;
    struct {
        short x;
        short y;
    } s;
    struct test *next;
  };

  

• Example 3:

Union

Data alignment

• Many computer systems place restrictions on the allowable addresses for the primitive data types, requiring that the address for some objects must be a multiple of some valueK (typically 2, 4, or 8). Such alignment restrictionssimplify the design of the hardware forming the interface between the processor and the memory system.

Example 1

struct S1 {
  int i;
  char c;
  int j;
}

• Suppose the compilter use the minimal 9-byte allocation, diagrammed as follows:

• Then it would be impossible to satisfy the 4-byte alignment requirement for both fields i (offset 0) and j (offset 5). Instead, the compiler inserts a 3-byte gap (shown here as shaded in blue) between fields c and j:

  As a result, j has offet 8, the overall structure size is 12 bytes

Example 2

  struct S2 {
    int i;
    int j;
    char c;
  }