Calculation of struct size and space optimization-Bit Field

/*************************************** ******************************* Author: samson * Date: 09/21/2013 * Test platform: * #1 SMP Debian 3.7.2-0 + kali8 * gcc (Debian 4.7.2-5) 4.7.2 ***************************************: http://blog.csdn.net/yygydjkthh/article/details/11850735 The length of a struct field with a bit field is calculated as follows, the length of the struct consisting of bitfield calculation and basic type values is not an algorithm. Test code and Result: # include <stdio. h> # include <stdlib. h> typedef struct BitArea1 {char c: 4; int l: 30;} BITAREA1; // The length of the above struct is 4bit + 30bit = 34bit, it is larger than a single byte (32bit), and because: 2nd of the alignment rules: the alignment value of the structure or class itself: the value with the largest alignment value of its members. Therefore, it should be aligned according to the four bytes of the int type. Therefore, the 34bit complement alignment occupies 8 bytes. If the sum of the two fields is less than or equal to 32bit, the space occupied by this struct is 4 Bytes: typedef struct BitArea1 {char c: 4; int l: 20;} BITAREA1; likewise, if the int type is modified to the short type, when the number of bits added to the two fields is smaller than or equal to sizeof (short) = 2 Byte = 16bit, the occupied space should be 2 bytes: typedef struct BitArea1 {char c: 4; short l: 10;} BITAREA1; if the number of bits added to the two fields is greater than 2 Byte = 16 bit, the space occupied is 4 bytes; typedef struct BitArea1 {char c: 4; short l: 13;} BITAREA1; typedef struct BitArea {Int a: 1; char B: 1; int c; char d: 3; char e: 5; int f; char g: 3;} BITAREA; // The length of the above struct should be 2 bits for a and B, and because a is of the int type and C is of the int type, according to the first and second rules, then a + B + c = 4 + 4 = 8, d + e is exactly 8 bit, And because f is int, d + e + f = 4 + 4 = 8, g is 3bit, 8 + 8 + 1 = 17, and according to the first rule: the alignment value of the struct or class: the value of the biggest alignment value of its members. The final result is 20; typedef struct BitAreaBetter {char B: 1; char d: 3; char e: 5; char g: 3; int c; int f; int: 1;} BITAREABETTER; // The length of the above struct should be 4 + 4 + 4 + 4 = 16 according to the previous conclusion; because C of the first four char Types is int type, the length of the two bytes added up by the first four char types should be supplemented by 4 bytes; typedef struct BitAreaBest {char B: 1; char d: 3; char e: 5; char g: 3; int a: 1; int c; int f;} BITAREABEST; // according to the previous summary, the length of the above struct can also be 4 + 4 + 4 = 12; the total space occupied by the first four char Types plus an int bit is smaller than the size of an int type. Therefore, the total space occupied is 4 bytes in the first rule, the last two values of the int type are 4 bytes. // The Optimization Method for the struct with bits field is: Put the bits field into one partition as much as possible; typedef struct BitAreaBest1 {int a: 1; int c; int f; char B: 1; char d: 3; char e: 5; char g: 3;} BITAREABEST1; // The length of the above struct is actually the same as that of BITAREABETTER, but the sequence is adjusted. int main () {int len, lenBest, lenBett, len01, lenBest01; BITAREA BitArea; BITAREA1 BitArea1; commandid; BITAREABEST BitAreaBest; commandid; len = sizeof (BitArea); lenBett = sizeof (bytes); lenBest = sizeof (BitAreaBest); len01 = sizeof (BitArea1 ); lenBest01 = sizeof (BitAreaBest01); printf ("BitArea1 len is % d BITAREABETTER len is % d BitAreaBest len is % d len01 is % d BitAreaBest01 len is % d \ n", len, lenBett, lenBest, len01, lenBest01); return 0;} root @ quieter:/usr/local/samsonfile/yygytest #. /. out BitArea1 len is 20 BITAREABETTER len is 16 BitAreaBest len is 12 len01 is 8 BitAreaBest01 len is 16