c++ - memory alignment- total size of structure multiple of structure alignement and not processing size -

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In the previous post, I understand why we should take alignment for a structure similar to the largest feature size.

Now, I would like to know that once we have chosen this alignment, we have to do padding so that the size of the total structure is of a majority of alignment structure and not the size of the processor word.

Here's an example: 

  #include & lt; Stdio.h & gt; // Alignment requirements // (Typical 32 bit machine) // four 1 byte // less int 2 bytes // init 4 bytes // double 8 bytes typedef struct structc_tag {char c; Double D; Ints; } Structc_t; Int main () {printf ("sizeof (structd_t) =% d \ n", size (structd_t)); Return 0; }   
 We can think that the size of structd_t is equal to 20 bytes:  
  char c; Four padding 1 [7] double D; Ints;   
 Because we've got a structure alignment equal to 8 (double D).  
 But, in fact, the total size is equal to 24 because 20 is not a value of 8 and we have to do padding after "int s" (four padding 2 [4]).  
 If I take an array of structure, then the first element of each structure is a good ad for the 32 bit processor (0, 20, 40) because the size of the processing word 20, 40 ... 4 Are there.  
 So why did I have to paddle for a multi of 8 (structure alignment), in this example 24 bits, instead of having 20 bytes (which is good for 32bit processor (word size = 4 bytes) Advertise: 0, 20, 40 ...?  
 Thanks for your help   
 
 
  Consider the struct_t array [n] . If the size of the structure is 20 (not many of the eight), the second element was aligned to the 4-byte border.  < P> To clarify:  array [0]  Address to  0 . If the size of the structure is e  20 , the  array [1]  starts with the address  20 , And this is the  d  address at land  28 , which is not a proper alignment for a double.

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