A summary of the size of bytes in VS

Description: With regard to byte alignment, many people gave the answer by setting the #pragma pack (1) to operate. However, in situations where space usage is not a concern, or if the protocol package is not being sent, this setting is not performed in order for the program to run faster.

The following are the default byte alignments

1. In a struct, the number of bytes used in the alignment is the maximum amount of bytes that are searched for keywords in the struct (the default minimum is 4 bytes)

Like what:

struct Node {

int A;

Double b;

};

The size of sizeof (Node) is 16 bytes instead of the so-called 12 bytes, because the number of bytes in the struct keyword (double) is 8 bytes, so the size is byte-aligned.

If this is the case:

struct Node {

int A;

Double b;

char c;

}

Then the size of sizeof (Node) is 24 bytes.

2, regarding the member function pointer occupies the space size, the concrete can see the following link

Http://www.linuxeden.com/html/news/20141206/157482.html

Note that if this is the case:

Class Base;

typedef (BASE::* Func) ();

Then func takes up 16 bytes

If

Class Base;

Class base{};

typedef (BASE::* Func) ();

Then Func takes up only four bytes. This means that as long as the class Base is defined in a typedef (BASE::* Func) (), it will simply take up a simple corresponding number of bytes.

3. What happens if the member function pointers are also put into the structure?

Declare a class here

Class Base;

typedef (BASE::* Func) ();

Through the 2nd link we know here the size of sizeof (Func) is 16 bytes

So here's an example

struct Node {

int A;

func func;

Double b;

}

How much space does sizeof need to occupy? The answer is 8+16+8 = 32 bytes. Since the structure is calculated with the largest footprint, why is it 32 bytes instead of 48 bytes? In fact, the alignment is the keyword (double) that is defined by C + +, as I said above. So it takes only 32 bytes.

A summary of the size of bytes in VS