# Pragma pack (N) and sizeof evaluate the structure size

To increase the CPU storage speed, VC performs "alignment" on the starting addresses of some variables. By default, VC specifies that the offset of the starting address of each member variable to the starting address of the structure must be a multiple of the bytes occupied by the variable type. Common alignment types are listed below (vc6.0, 32-bit system ).

Type
Alignment (offset of the starting address of the variable to the starting address of the structure)

Char
The offset must be a multiple of sizeof (char), that is, 1.

Int
The offset must be a multiple of sizeof (INT), that is, 4.

Float
The offset must be a multiple of sizeof (float), that is, 4.

Double
The offset must be a multiple of sizeof (double), that is, 8.

Short
The offset must be a multiple of sizeof (short), that is, 2.

When each member variable is stored, the space is requested in sequence based on the order in which the structure appears, and the positions are adjusted according to the alignment above. The vacant byte VC is automatically filled. At the same time, to ensure that the size of the structure is a multiple of the number of byte boundary values of the structure (that is, the number of bytes occupied by the Type occupying the maximum space in the structure, therefore, after applying for space for the last member variable, the vacant bytes will be automatically filled as needed.

 

Generally, you can use the following method to change the default peer condition:
· With the pseudo command # pragma pack (n), the C compiler will align according to n Bytes.
· Use the pseudo command # pragma pack () to cancel the custom byte alignment.

In addition, the following method is provided:
· _ Attribute (aligned (N) to align the structure members to the natural boundary of n Bytes. If the length of a member in the structure is greater than N, the maximum member length is used for alignment.
· _ Attribute _ (packed): cancels the optimization alignment of the structure during compilation and alignment according to the actual number of bytes occupied.

The first method above n = 1, 2, 4, 8, 16... is more common.

Struct mystruct

{

Double dda1;

Char DDA;

Int type

};

When allocating space for the above structure, VC allocates space for the first member dda1 according to the sequence and alignment of the member variables, the starting address is the same as the starting address of the structure (the offset 0 is just a multiple of sizeof (double). The member variable occupies eight bytes; next, allocate space for the second member DDA. the offset of the next address that can be allocated to the starting address of the structure is 8, which is a multiple of sizeof (char, therefore, the DDA is stored in an alignment where the offset is 8. This member variable occupies sizeof (char) = 1 byte, and then allocates space for the third member type, in this case, the offset of the next allocable address to the starting address of the structure is 9, not a multiple of sizeof (INT) = 4. To meet the offset constraints of alignment, VC automatically fills in three bytes (these three bytes do not have anything). At this time, the next address that can be allocated has a 12 offset to the starting address of the structure, which is exactly sizeof (INT) = a multiple of 4, so the type is stored in the offset Where the value is 12, the member variable occupies sizeof (INT) = 4 bytes. At this time, the member variables in the entire structure have been allocated space, and the total occupied space is: 8 + 1 + 3 + 4 = 16, which is exactly the number of bytes in the structure (that is, the number of bytes occupied by the Type occupying the maximum space in the structure sizeof (double) = 8) so there is no vacant byte to fill. Therefore, the size of the entire structure is sizeof (mystruct) = 8 + 1 + 3 + 4 = 16. Among them, three bytes are automatically filled by VC and nothing makes sense.

Next, let's take another example to change the position of the member variable of mystruct above to the following:

Struct mystruct

{

Char DDA;

Double dda1;

Int type

};

How much space does this structure occupy? In the vc6.0 environment, we can obtain that sizeof (mystruc) is 24. Based on the space allocation principles mentioned above, we will analyze how VC allocates space for the above structure.
(Simple description)

Struct mystruct

{

Char DDA; // The offset is 0. The alignment is satisfied and the DDA occupies 1 byte;

Double dda1; // the offset of the next available address is 1, not sizeof (double) = 8

// Multiple. You need to add 7 bytes to make the offset 8 (the alignment is satisfied ).

// Method). Therefore, VC automatically fills in 7 bytes, and dda1 is stored at an offset of 8

// Address, which occupies 8 bytes.

Int type; // the offset of the next available address is 16, which is twice that of sizeof (INT) = 4.

// Number, which meets the alignment of int, so no VC auto-fill is required.

// Put the address at the offset of 16, which occupies 4 bytes.

}; // All member variables are allocated space. The total size of the space is 1 + 7 + 8 + 4 = 20, not a structure.

// Number of knots (that is, the number of bytes occupied by the largest space type in the structure sizeof

// (Double) = 8), so four bytes need to be filled to meet the structure size

// Sizeof (double) = a multiple of 8.

Therefore, the total size of this structure is: sizeof (mystruc) is 1 + 7 + 8 + 4 + 4 = 24. Among them, 7 + 4 = 11 bytes are automatically filled by VC, and nothing makes sense.

The special processing of the structure storage by VC does increase the speed of the CPU storage variable, but sometimes it also brings some trouble. We also shield the default alignment of the variables, you can set the alignment of variables. # Pragma pack (n) is provided in VC to set the variable to n-byte alignment. N-byte alignment means the offset of the Start address of the variable:
First, if n is greater than or equal to the number of bytes occupied by the variable, the offset must meet the default alignment,
Second, if n is less than the number of bytes occupied by the variable type, the offset is a multiple of N, and the default alignment is not required.
The total size of the structure also has a constraint, which is divided into the following two cases: if n is greater than the number of bytes occupied by all member variable types, the total size of the structure must be a multiple of the space occupied by the largest variable; otherwise, it must be a multiple of N. The following is an example of its usage.

# Pragma pack (push) // save alignment status

# Pragma pack (4) // set to 4-byte alignment

Struct Test

{

Char M1;

Double M4;

Int m3;

};

# Pragma pack (POP) // restore alignment

The size of the above structure is 16. Next we will analyze its storage situation. First, we will allocate space for M1, and its offset is 0, which meets our own alignment (4-byte alignment ), m1 occupies 1 byte.
Then we start to allocate space for M4. At this time, the offset is 1 and three bytes need to be supplemented. In this way, the offset must be a multiple of N = 4 (because sizeof (double) is greater than N ), m4 occupies 8 bytes.
Then allocate space for M3. At this time, the offset is 12, which must be a multiple of 4. m3 occupies 4 bytes. At this time, space has been allocated for all member variables. A total of 16 bytes are allocated, which is a multiple of N.
If you change # pragma pack (4) to # pragma pack (16), the size of the structure is 24. (Please analyze it by yourself)

 

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# Pragma pack (8)

Struct S1

{
Short;
Long B;
};

Struct S2

{
Char C;
S1 D;
Long long E;
};

# Pragma pack ()

Question
1. sizeof (S2) =?
2. How many bytes are left behind C in S2 and then D?

 

The result is as follows:

The sizeof (S2) result is 24.
An important condition for member alignment is that each member is aligned separately. That is, each member is aligned in its own way.
That is to say, although alignment by 8 bytes is specified above, not all Members are alignment by 8 bytes. the alignment rule is that each member has a smaller alignment according to its type alignment parameter (usually the size of this type) and the specified alignment parameter (8 bytes here. in addition, the length of the structure must be an integer multiple of all alignment parameters used.
In S1, member A is 2 bytes aligned by 2 bytes by default, and the specified alignment parameter is 8. Among the two values, 1 and a are aligned by 1 bytes. Member B is 4 bytes, the default value is 4-byte alignment. In this case, it is 4-byte alignment, so sizeof (S1) should be 8;
In S2, C is the same as a in S1. It is aligned by 2 bytes, while D is a structure. It is 8 bytes. What is its alignment? For the structure, its default alignment is the largest of all its members using alignment parameters, and S1 is 4. therefore, member D is aligned in 4 bytes. the member e is 8 bytes, which is aligned by 8 bytes by default. It is the same as the specified one, so it is connected to the boundary of 8 bytes. At this time, 12 bytes are used, therefore, four bytes are added, and the member e is placed starting from 16th bytes. at this time, the length is 24 and can be divisible by 8 (member e is aligned by 8 bytes. in this way, a total of 24 bytes are used.
A B
Memory layout of S1: 11 **, 1111,
C s1.a s1. B d
Memory layout of S2: 1 ***, 11 ***, 1111, *** 11111111

There are three important points:
1. Each Member is aligned in its own way, and the length can be minimized.
2. The default alignment of complex types (such as structures) is the alignment of its longest member, so that the length can be minimized when the member is a complex type.
3. The alignment length must be an integer multiple of the largest alignment parameter in the member. This ensures that each item is bounded when processing arrays.