Thread: memcpy question

I'm seeing a weird problem with memcpy. I have the following code:

Code:

int main()
{
        char data[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 };


        struct mystruct {

                char a;
                short b;
                char c;
                long d;
                char e;
                short f;
                char g;
        } ms;

        memcpy(&ms, data, 16);

Ok, I expected a = 1, but for some bizarre reason, c = 5! It skipped over 4 completely. Not only that, but b = 3 and 4, it skipped over 2! This totally messes up all of my code.

Whats the best way to fix this issue without manually specifying offsets (i.e. c = data[3]) and make sure all the other values equal what they are supposed to?

Google struct padding.
Even if there's no padding, it won't work.(as what you expect).
Just assign to individual member.

Originally Posted by Bayint Naung

Google struct padding.
Just assign to individual member.

I know its padding the structure. My question is, how do I solve that problem without manually specifying offsets? It becomes an issue with portability and endianness between machines, and much further complicates the code.

Structs may include padding (spare space between members) for performance reasons. memcpy() will happily write data to those padding bytes.

Rather than trying to use memcpy() to initialise the struct - such things are sensitive to how the compiler lays out the structure in memory - initialise the struct directly.

Code:

struct mystruct ms = {1, 1027, 4, <other values you expect>};

Code:

ms.a = data[0];
ms.b = (data[1]<<8) + data[2];

etc etc
If you want portability across all endian implementations and struct packing.

Reorder the variables in your struct so that they are ordered from largest to smallest. (long -> short -> short -> char -> char -> char -> char)
That minimises the padding.