CODEWITHSUNDEEP
c++cmemory-alignment
user1773603
user1773603

Reputation:

memory alignment- total size of structure multiple of structure alignement and not processing size

In a previous post, I have understood why we must take an alignment for a structure equal to the biggest attribute size.

Now, I would like to know why, once we have chosen this alignment, we have to do padding so that the total structure size is a multiple of the structure alignement and not the processor word size.

Here an example :

#include <stdio.h>

// Alignment requirements
// (typical 32 bit machine)

// char         1 byte
// short int    2 bytes
// int          4 bytes
// double       8 bytes

typedef struct structc_tag
{
   char        c;
   double      d;
   int         s;
} structc_t;

int main()
{
   printf("sizeof(structd_t) = %d\n", sizeof(structd_t));

   return 0;
}

We could think that size of structd_t is equal to 20 bytes with :

   char        c;
   char Padding1[7]
   double      d;
   int         s;

because we have taken a structure alignement equal to 8 (double d).

But actually, total size is equal to 24 because 20 is not a multiple of 8 and we have to do padding after "int s" (char Padding2[4]).

If I take an array of structure, the first elements of each structure is at good adresses for a 32 bits processor (0, 20, 40) because 20, 40 ... are multiple of 4 (processing word size).

So Why have I got to do padding for having a multiple of 8 (the structure alignement), i.e 24 bytes in this example, instead of having 20 bytes (which gives good adresses for the 32 bits processor (word size = 4 bytes) : 0, 20, 40 ... ?

Thanks for your help

Upvotes: 1

Views: 583

Answers (1)

user58697
user58697

Reputation: 7923

Consider struct_t array[n]. Would the size of structure be 20 (not multiple of 8), the second element were aligned at a 4-byte boundary.

To clarify: Let array[0] be at address 0. If size of the structure is 20, array[1] starts at address 20, and it's d lands at address 28, which is not a proper alignment for a double.

Upvotes: 3

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