Question

I’m writing C cross-platform library but eventually I’ve got error in my unittests, but only on Windows machines. I’ve tracked the problem and found it’s related to alignment of structures (I’m using arrays of structures to hold data for multiple similar objects). The problem is: memset(sizeof(struct)) and setting structures members one by one produce different byte-to-byte result and therefore memcmp() returns ‘not equal’ result.

Here the code for illustration:

#include <stdio.h> #include <string.h>  typedef struct {     long long      a;     int            b; } S1;  typedef struct {     long           a;     int            b; } S2;  S1 s1, s2;  int main() {     printf('%d %d\n', sizeof(S1), sizeof(S2));      memset(&s1, 0xFF, sizeof(S1));     memset(&s2, 0x00, sizeof(S1));      s1.a = 0LL; s1.b = 0;      if (0 == memcmp(&s1, &s2, sizeof(S1)))         printf('Equal\n');     else         printf('Not equal\n');      return 0; } 

This code with MSVC 2003 @ Windows produce following output:

16 8 Not equal 

But the same code with GCC 3.3.6 @ Linux works as expected:

12 8 Equal 

This makes my unit-testing very hard.

Am I understand correctly that MSVC uses size of biggest native type (long long) to determine alignment to structure?

Can somebody give me advice how can I change my code to make it more robust against this strange alignment problem? In my real code I’m working with arrays of structures via generic pointers to execute memset/memcmp and I’m usually don’t know exact type, I have only sizeof(struct) value.

Answer 1

What we have done is used the #pragma pack to specify how big the objects should be:

#pragma pack(push, 2)  typedef struct {     long long      a;     int            b; } S1;  typedef struct {     long           a;     int            b; } S2;  #pragma pack(pop) 

If you do this, the structures will be the same size on both platforms.