&gt; **Possible Duplicate:**  
&gt; [Why isn&amp;rsquo;t sizeof for a struct equal to the sum of sizeof of each member?](https://stackoverflow.com/questions/119123/why-isnt-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member)  


Consider the following C code:

    #include &lt;stdio.h&gt;    
    
    struct employee
    {
      int id;
      char name[30];  
    };

    int main()
    {
      struct employee e1;      
      printf(&quot;%d %d %d&quot;, sizeof(e1.id), sizeof(e1.name), sizeof(e1));
      return(0);
    }

The output is:

4 30 36

Why is the size of the structure not equal to the sum of the sizes of its individual component variables?



size of struct in C

I&#39;m working on a 32-bit machine, so I suppose that the memory alignment should be 4 bytes. Say I have this struct:

    typedef struct {
    	unsigned short v1;
    	unsigned short v2;
    	unsigned short v3;
    } myStruct;

The plain added size is 6 bytes, and I suppose that the aligned size should be 8, but `sizeof(myStruct)` returns me 6.

However if I write:

    typedef struct {
    	unsigned short v1;
    	unsigned short v2;
    	unsigned short v3;
    	int i;
    } myStruct;

the plain added size is 10 bytes, aligned size shall be 12, and this time `sizeof(myStruct) == 12`.

Can somebody explain what is the difference?

All Structure Packing Solutions on Gang of Coders

Total of 2 Structure Packing Solutions

size of struct in C

Memory alignment in C-structs