How are the structure members stored on a little endian machine?

struct Dummy {
  int x;
  char y;
};

int main() {
  struct Dummy dum;
  dum.x = 10;
  dum.y = 'a';
}

How would be the layout o开发者_如何学Gof the structure members on a little endian machine?

Would it be something like this?

  0x1000  +0   +1   +2   +3
         ___________________
    x:  | 10 |  0 |  0 |  0 |     
         -------------------       
    y:  | 'a'|  0 |  0 |  0 |
         -------------------
  0x1000  +4   +5   +6   +7 

I think you'll find this question useful. The endianess is usually relevant for a word in the memory, not to the whole structure.

Structure layout is a compiler implementation detail, affected by the default packing. Endianness normally only affects the order of the bytes in a structure member value, not the layout. Check the small print in the compiler manual or use sizeof and the offsetof macro to experiment.

The layout you documented in your question is indeed very common for a 32-bit LE compiler.

The structure members will be in the order declared, with padding inserted as necessary so each field is properly aligned for its type and with padding inserted as necessary at the end such that in an array each subsequent structure is properly aligned and begins immediately after the end of the previous structure. It is also possible (but unlikely) that additional unnecessary padding will be inserted between any two elements or at the end.

Each field itself will be stored as appropriate for the type on the compiler and architecture, e.g. the int 10 would be stored as the bytes 0a 00 00 00 on a normal little-endian machine with 32-bit ints.