Find the memory occupied by the structure

Find the space occupied by a certain type , We need to use sizeof keyword

for example ：sizeof(int)

           sizeof(char)

           sizeof(double)

So if sizeof Inside the bracket is a structure ？ A structure is a collection of many types , How can we find the space it occupies ？

There is a point of knowledge here ： Structure memory alignment

1. first Members are related to structural variables The offset for the 0 The address of

2. Other member variables should be aligned to a An integer multiple of the alignment number The address of

Align numbers = Compiler default alignment number And The size of the member Smaller value

vs The default value in is 8

Set the default number of alignments ：

#pragma pack(6) // Set the default alignment number to 6

#pragma pack(0)//0 Invalid value , The minimum is 1

3. structure Total body size yes Maximum number of alignments （ Each member variable has an alignment number ） Of An integral multiple .

4. If the structure is nested , The nested structure is aligned to an integral multiple of its maximum alignment , The overall size of the structure is the maximum number of alignments （ The number of alignments with nested structures ） Integer multiple

Example ：

struct S1

{

        char c1;

        int i;

        char c2;

};

sizeof(struct S1); --- 12

Why is there memory alignment ？

1. Platform reasons （ Reasons for transplantation ）

Not all hardware platforms can access any data on any address ;

Some hardware platforms can only access certain types of data at certain addresses , Otherwise, a hardware exception will be thrown

2. Performance reasons ：

data structure （ Especially stacks ） It should be aligned on the natural boundary as much as possible ,

The reason lies in , To access unaligned memory , The processor needs to do two memory accesses ;

The aligned memory access only needs one access

3. On the whole ：

The memory alignment of the structure is Space In exchange for Time How to do it

How to save space , It saves time ？

The answer is ： Let the members who occupy less space gather together as much as possible

for example , For the example above , We make the following modifications

struct S1

{

        char c1;

        char c2;

        int i

};