How Does Data Alignment Affect the Size of C Classes?

Understanding Class Size Determination in C

Determining the size of a C class at compile-time requires an understanding of data alignment principles. Each member within a class has a size and alignment requirement.

Size and Alignment Calculation Process

During compilation, the compiler initializes a running size (S) to zero and an alignment requirement (A) to one (byte). For each member in the class:

1. The alignment requirement (a) is checked. If S is not a multiple of a, S is adjusted to the nearest multiple. This determines the offset of the member.
2. The least common multiple (LCM) of A and a is calculated, and A is updated to LCM.
3. Space (s) for the member is added to S.

After processing all members, the final size of the class is determined. It is the value of S adjusted to be a multiple of the alignment requirement of the entire class (A).

Alignment Considerations

• Arrays: Size is the number of elements multiplied by element size, and the alignment requirement is the same as an element.
• Structures: Size and alignment requirements are determined recursively.
• Unions: Size is the size of the largest member, rounded up to the LCM of all members' alignments.

Example

The provided code demonstrates this process:

<code class="cpp">#include <xmmintrin.h>

class TestClass1 {
  __m128i vect;
}; // Size: 16 bytes

class TestClass2 {
  char buf[8];
  char buf2[8];
}; // Size: 16 bytes

class TestClass3 {
  char buf[8];
  __m128i vect;
  char buf2[8];
}; // Size: 48 bytes

class TestClass4 {
  char buf[8];
  char buf2[8];
  __m128i vect;
}; // Size: 32 bytes</code>

TestClass3, despite having the same members as TestClass1 and TestClass2, is larger (48 bytes) due to the alignment requirement of __m128i, which forces a 16-byte boundary. TestClass4, with a different data member ordering, has a different alignment and size (32 bytes).

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