Sizeof Structure as Defined, Not Allocated

Sizeof Structure as Defined, Not Allocated

Question:
Given:

pre>typedef struct{ byte a; byte b; byte c; byte d; byte e;} LETTERS;

sizeof(LETTERS) returns eight bytes because of padding by the compiler for alignment.How can I determine how many bytes were actually defined, not allocated?

Answer:
There is no standard function or operator to detect how many padding bytes an aggregate contains. However, can use the following expression to calculate the number of padding bytes:

sizeof(LETTERS)-(n*sizeof(byte))

Where n is the number of byte members in LETTERS.

Most compilers have special flags that control aggregate member alignment. On 32-bit platforms a typical alignment boundary is four bytes. You can increase it or reduce it. By setting it to one, your struct will contain no padding bytes at all. However, remember that this can incur significant runtime overhead.

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