Structure knowledge points all in

Structure ：

Definition ：

        struct Type name

        {

                （ Member list \ attribute ）

                 member 1;

                 member 2;

                .....

        };  // The key is ： What information is needed

        eg:

        struct Student

        {

                char name[20];

                int age;

        };// Definition complete , Its function is the same as that of the built-in type , Type design does not take up memory space

Use ：

         Define variables by type , Then initialize the data in the variable one by one according to the definition order

        eg:

        struct Student student = { " Li Si ",14};

        struct Student student1;// If it's a local variable ( It's written inside the function )-> uninitialized , The member contents are random values , Error will be reported in compilation ; If it's a global variable ( Write outside function )-> uninitialized , The compiler initializes the member list by default int,short Those are 0,char The type is \0, Compile without error

         Example solution ：

            struct Student a;// utilize struct Student The data type defines a variable a;

            struct Student b = { " Cao Cao ",20};

            b.age = 21;// The common variables of the structure are passed through “.” Visit its members

            (1) modify b My name is Liu Bei

             error ：b.name=” Liu bei ”// String assignment to array... Is not allowed , You should use a loop , It has nothing to do with the structure

            correct ：strcpy(b.name, " Liu bei "); 

            (2) Output b The data of ： Name and age

            printf(" % s, % d", b.name, b.age);

            (3) take a Change it to “ king of Wu in the Three Kingdoms Era ”,18

            strcpy(a.name, " king of Wu in the Three Kingdoms Era ");

                a.age = 18;

Simplify the use ：

        1.struct Type name

          { } Alias ;

        2.typedef struct Type name   Alias ;

Nested structure ：

         It can be used in the structure Previously stated Type of structure

        eg:

        struct Date

        {

                int year;

                int month;

                int day;

        }Date;

        struct Student

        {

                const char* name;

                 int age;

                Date birthday;

        };

        int main()

        {

                struct Student s1 = { " Li Si ",14,{2001.9.18}};

        }

size ：

        test ：printf(“%d”,sizeof(struct Student));

        computing method ： Related to memory alignment , The following is written .

Array of structs ：

        eg:struct Studeng arr[]={ {},{}...};

visit ：

        arr[i].name,arr++

Member visit ：

         Variable name . Structural variable   ‘.’ Member accessor

        eg:student.name

How to understand byte alignment problem ：

1. The basic unit of memory is bytes , In theory , Variables can be accessed at any address , But actually CPU It does not read memory byte by byte , But rather 2,4,8 Read and write multiple of memory block , Therefore, the address of basic data type is limited , Then it requires that various data types be arranged in space according to certain rules .
2. Some platforms start at my address every time , And if it's stored at the beginning of the odd address , Need 2 All read cycles , And the high and low bytes of the result read out twice can be pieced together to get the 32bit data . Obviously, there is a lot of decrease in reading efficiency .
3. Due to different platforms, alignment may be different . In this way , The same structure may have different sizes on different platforms , Unconsciously , The data sent to each other may be out of order , It even caused serious problems .

How to calculate the size of a structure ：

1. The first address of the structure variable address , Must be in a struct variable “ The maximum number of bytes occupied by basic data type members ” Integer multiple .

        eg:struct Student

        {

        char name[20]; //1*20=20 byte

        int age; //4 byte

        };

         because 20 Is the largest byte in the structure ,so The first address of a struct variable must be 20 Multiple . Pictured

       -> The multiple is 0

2. The offset of each member in the structure variable from the first address of the structure , Are integer multiples of the bytes occupied by the basic data type of the member .

3. The total size of the structure variable , Is in the structure variable “ Maximum Basic data type Number of bytes occupied by members ” The integer of

times （ Be careful ：struct And array are not basic data types ）

Give an example of ： Allocate the bytes shown in the figure in memory

（1）

Explain ：char cha:1 byte ,char chb:1 byte ,int ia：4 byte , At least 4 Position of , So two bytes are wasted .1+1+2+4=8 yes 4 Multiple , Therefore, the byte size of the structure is 8

（2）

  Explain ：cha:1,da:8 Need to waste in front 7 Bytes ,chb:1.8+8+1=17. Minimum 24.

 

（3）

Explain ：ch:3,ia:4, Waste one in front ,4+4=8, yes 4 Multiple , Then for 8.

 

 

（4） Nested structure analysis ：

Explain ：10+8+12=30,4*8=32 waste 2 Bytes .

32+8=40, The basic data type is double:8 byte ,40 yes 8 Multiple , The structure byte size is 40.

Instruction aligner ：

Preprocessing instruction #pragma pack(n), You can change the default number ,n The values for 1,2,4,8,16.

If #pramga pack(n) Medium n Larger than the number of bytes occupied by any one of the structure members , Then n Invalid value . The compiler will select the number of bytes of the largest data member in the structure as the benchmark for its .

eg: The number of bytes is 2,a Read 1 Time , waste 1 Space ,b Read twice ,c Read 1 Time , waste 1 Space .

  The number of bytes is 4,a Read 1 Time , waste 3 Space ,b Read 1 Time ,c Read 1 Time , waste 3 Space .

 

Structure pointer ：

Definition ：

        typedef struct Student* stu;

         perhaps

        struct Student

        {

                char name[20];

                int age;

        }*stu;

Use ：

         Define pointer variables by type , And then give it a direction .

        eg:

        Student s1 = { " Zhang San ",20 };

        typedef struct Student* stu;

        stu S=&s1;

         adopt S Output s1 The content of .

        eg:

        printf(“%c,%d”,S->name,S->age);

Member visit ：

1.(* Variable name ). Structural variable  ‘.’ Member accessor

(*S).age; Must have parentheses , because “.” Has a higher priority than “*”

2. Variable name -> Structural variable    ‘->’ Pointer

S->age;