Understanding offsetof(struct ,member);

Code:

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#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
int main(void)
{   
struct s
{        int i;
        char c; 
      double d; 
      char a[]; 
  };  /* Output is compiler dependent */

  printf("offsets: i=%ld; c=%ld; d=%ld a=%ld\n",(long) offsetof(struct s, i), (long) offsetof(struct s, c), (long) offsetof(struct s, d), (long) offsetof(struct s, a)); 
 printf("sizeof(struct s)=%ld\n", (long) sizeof(struct s));
  exit(EXIT_SUCCESS);
}

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Output : offsets: i=0; c=4; d=8 a=16sizeof(struct s)=16
My Hypothesis :
Integer is 4 bytes ..
its first index so i=0;
Character is 1 byte ..
but d=8 it should be 5.Can someone explain me how offset works..

Quote:

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Originally Posted by thriller500

Character is 1 byte ..but d=8 it should be 5.

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First off, I don't know why all your code is on one line, but if you want anyone to read it, make sure it's formatted correctly.

offsetof() works as you think - it gives the offset, in bytes, of the given member of the structure. However, in a struct the compiler is allowed to lay out variables however it wants, including reordering variables and adding "padding" or unused bytes between members. This is usually to get good/needed alignment to make faster code. Basically, just trust what offsetof() says.

> including reordering variables
This isn't true at all.

Quote:

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Originally Posted by C99

Semantics
4 As discussed in 6.2.5, a structure is a type consisting of a sequence of members, whose
storage is allocated in an ordered sequence, and a union is a type consisting of a sequence
of members whose storage overlap.

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The size of each member, and the padding between them is compiler dependent, but the order itself is fixed in the order written in the source code.

Otherwise, you would never be able to initialise a struct if the compiler were free to rearrange the order.

I didn't know that. Although, to be pedantic:

Quote:

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Originally Posted by C99

Semantics
4 As discussed in 6.2.5, a structure is a type consisting of a sequence of members, whose storage is allocated in an ordered sequence, and a union is a type consisting of a sequence of members whose storage overlap.

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Just because it's an ordered sequence doesn't imply its members have the same order in memory as in the definition - however, I see that a later point does state that this must be the case.

Quote:

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Originally Posted by JohnGraham

Just because it's an ordered sequence doesn't imply its members have the same order in memory as in the definition - however, I see that a later point does state that this must be the case.

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The reason the standard defines an order of members is that, before such a thing was standardised, compilers chose a defacto order and a lot of code came to rely on that order.