On the algorithm of the structure byte of C + +

Simple answers, see examples.

I machine 64 bits, pointer 8 bytes; 32 bits, pointer 4 bytes.

Files: SIZEOF.C

#include  <stdio.h>typedef struct SS{    char     name[131];    short   type;               int     size;               int     offset;             unsigned short      Precision;                void     *data;                            #if  0}__attribute__  (( __packed__) AA; #else}colattr_t; #endifint  main (INT&NBSP;ARGC,&NBSP;CHAR&NBSP;**ARGV) {    printf ("%d \n",  sIzeof (AA));     return 0;} 

Seemingly 151 bytes (#if 1 can be seen, attribute compact byte network), the actual 160 bytes.

Answer: is 8-byte aligned, because void * is 8 bytes, minus void * Maximum is 4 bytes, which is aligned by 4 bytes .

650) this.width=650; "src=" Http://s3.51cto.com/wyfs02/M01/71/41/wKioL1XJ2biSLEeWAABBh1IyLyA483.jpg "title=" Boke.jpg "alt=" Wkiol1xj2bisleewaabbh1iylya483.jpg "/>

Analysis: All are aligned to 8 bytes.

The first line, Char cname[128] bytes.

The second line, with 3 bytes and a short type, is 8 bytes in combination. (A total of 5 bytes, more than 3 of the int type).

In the third row, the int size and int offset combinations account for 8 bytes.

Row four, void * Because 64 bits account for 8 bytes.

This article is from the "Self-summary" blog, please be sure to keep this source http://2714254.blog.51cto.com/2704254/1683780

On the algorithm of the structure byte of C + +