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Most processors require specific memory alignment on variables certain types. Normally the minimum alignment is the size of the basic type in question, fo instance this is common

char variables can be byte aligned and appear at any byte boundary

short (2 byte) variables must be 2 byte aligned, they can appear at any even byte boundary. This means that 0x10004567 is not a valid location for a short variable but 0x10004566 is.

long (4 byte) variables must be 4 byte aligned, they can only appear at byte boundarys that are a multiple of 4 bytes. This means that 0x10004566 is not a valid location for a long variable but 0x10004568 is.

Structure padding occurs because the members of the structure must appear at the correect byte boundary, to achieve this the compiler puts in padding bytes (or bits if bit fields are in use) so that the structure members appear in the correct location. Additionally the size of the structure must be such that in an array of the structures all the structures are correctly aligned in memory so there may be padding bytes at the end of the structure too

struct example {
char c1;
short s1;
char c2;
long l1;
char c3;
}

In this structure, assuming the alignment scheme I have previously stated then

c1 can appear at any byte boundary, however s1 must appear at a 2 byte boundary so there is a padding byte between c1 and s1.

c2 can then appear in the available memory location, however l1 must be at a 4 byte boundary so there are 3 padding bytes between c2 and l1

c3 then appear in the available memory location, however because the structure contains a long member the structure must be 4 byte aligned and must be a multiple of 4 bytes in size. Therefore there are 3 padding bytes at the end of the structure. it would appear in memory in this order

c1
padding byte
s1 byte 1
s1 byte 2
c2
padding byte
padding byte
padding byte
l1 byte 1
l1 byte 2
l1 byte 3
l1 byte 4
c3
padding byte
padding byte
padding byte

The structure would be 16 bytes long.

re-written like this

struct example {
long l1;
short s1;
char c1;
char c2;
char c3;
}

Then l1 appears at the correct byte alignment, s1 will be correctly aligned so no need for padding between l1 and s1. c1, c2, c3 can appear at any location. The structure must be a multiple of 4 bytes in size since it contains a long so 3 padding bytes appear after c3

It appears in memory in the order

l1 byte 1
l1 byte 2
l1 byte 3
l1 byte 4
s1 byte 1
s1 byte 2
c1
c2
c3
padding byte
padding byte
padding byte

and is only 12 bytes long.

Originally Posted On : 2010-03-31 04:33:46

Anshul Makkar, anshul_makkar@justkernel.com

 

Tags: Structure Padding