Struct members ordering advantages in C

I know that in C, the order of the struct members is the order in which they are declared plus (usually , unless stated otherwise) padding might occur , leading to the two first structs I created with their respective size:

struct MyStructV1
{
    int a;
    char c;
    int b;  
};

printf("size of MyStructV1 is:%lu \n",sizeof(struct MyStructV1)); // prints 12

struct MyStructV1Packed
{
    int a;
    char c;
    int b;
}__attribute__((__packed__));;

printf("size of MyStructV1Packed is:%lu \n",sizeof(struct MyStructV1Packed)); // prints 9 

So far so good.

My question is, then, does the following declaration of "the unpacked" version of MyStructV2 has any benefits/advantages over the first version of MyStructV1 ?

struct MyStructV2
{
    int a;
    int b;
    char c;
};

printf("size of MyStructV2 is:%lu \n",sizeof(struct MyStructV2)); // prints 12 as well 

Note that now the order of the members has changes (b is declared before c).

I'm referring to memory access "costs" or cycles needs to be done in order to read/write the struct members and/or any other considerations that are relevant ?

Is it compiler/architecture dependent (if at all) ?

Cheers,

Guy.

There should be little or no difference between the first and last versions. In both structs, the int members are aligned on word boundaries, so they can be read from memory efficiently. The only difference between them is that the padding is between c and b in MyStructV1, and at the end of the structure in MyStructV2. But there is no execution overhead due to padding; accessing struct members is done simply by adding a known offset to the address of the beginning of the struct. In MyStructV1 the offset of b is 8, in MyStructV2 it's 4.