Determining the Size of a C Class During Compilation
Understanding Class Size and Memory Alignment
In C , class size is determined statically during compilation to optimize memory allocation and access. To calculate class size effectively, the compiler considers various factors, including data member sizes and alignment requirements.
Data Member Alignment and Size Calculation
For Plain Old Data (POD) classes with data members having deterministic sizes, the following rules generally apply:
- Each data member possesses a size (s) and an alignment requirement (a).
- The compiler's initial size (S) is set to 0, while alignment (A) is set to 1 byte.
-
Members are processed sequentially:
- Check the alignment requirement (a). If S is not divisible by a, increment S to align the member at the correct offset.
- Update A to the least common multiple of the current A and a.
- Increase S by s to accommodate the member.
- After processing all members, ensure S is divisible by A; if not, increase S accordingly.
- The final value of S represents the class size.
Additional Considerations
- Arrays: Size is calculated as the number of elements multiplied by the element size; alignment is the alignment requirement of an element.
- Structures: Size and alignment are calculated recursively using the same rules.
- Unions: Size is the size of the largest member plus any padding to meet the least common multiple of the alignments of all members.
Example: TestClass3
For TestClass3, the size calculation proceeds as follows:
- buf[8] requires 8 bytes with alignment 1, so S becomes 8.
- __m128i vect requires 16 bytes with alignment 16. S is first increased to 16 for alignment, then to 32 to accommodate vect.
- buf2[8] requires 8 bytes with alignment 1. S is increased to 24.
- As 24 is not divisible by 16, S is incremented by 8 to 32.
Thus, TestClass3 is 32 bytes in size.
Conclusion
By adhering to these alignment and size calculation rules, the compiler ensures optimal memory allocation and efficient data access for C classes.
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