calculating the size of structure

Question

I was going through the concept of structure member alignment and structure padding concept when I found that the size of this struct ...

struct nod         


        

Answer 1

Data structure alignment.

Described wonderfully here.

Suppose you have this structure:

struct  {
    char a[3];
    short int b;
    long int c;
    char d[3];
    };

Now, you might think that it ought to be possible to pack this structure into memory like this:

+-------+-------+-------+-------+
|           a           |   b   |
+-------+-------+-------+-------+
|   b   |           c           |
+-------+-------+-------+-------+
|   c   |           d           |
+-------+-------+-------+-------+

But it's much, much easier on the processor if the compiler arranges it like this:

+-------+-------+-------+
|           a           |
+-------+-------+-------+
|       b       |
+-------+-------+-------+-------+
|               c               |
+-------+-------+-------+-------+
|           d           |
+-------+-------+-------+

In the ''packed'' version, notice how it's at least a little bit hard for you and me to see how the b and c fields wrap around? In a nutshell, it's hard for the processor, too. Therefore, most compilers will ''pad'' the structure (as if with extra, invisible fields) like this:

+-------+-------+-------+-------+
|           a           | pad1  |
+-------+-------+-------+-------+
|       b       |     pad2      |
+-------+-------+-------+-------+
|               c               |
+-------+-------+-------+-------+
|           d           | pad3  |
+-------+-------+-------+-------+

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^{This post explained how padding is done, not why it is done. Please refer haccks' answer for why it is done.}