Structure alignment trap [plus]

Sometimes, the "obvious" things that have been paused for a long time in my mind are basically wrong. Let's take a look at the following points:
Struct t
{
Char ch;
Int I;
};
What kind of answer will I get when I use sizeof (t? If you do not want to think about it before, in 32-bit machines, Int Is 4 bytes, char is 1 byte, so T is a total of 5 bytes. The answer is actually 8 bytes, tested in vc6. Well, I am always hurt, and I am a little numb. Please accept it honestly! Why is the answer different from what you think? This introduces the concept of memory alignment.

Many real computer systems have limits on the locations where basic data is stored in the memory. They require that the first address value of the data be K (usually 4 or 8) this is the memory alignment, and this K is called the alignment modulus of the data type ). When the ratio of the alignment modulus of one type of S to the alignment modulus of another type of T is an integer greater than 1, we call it the alignment requirement of type s stronger than that of T (strict ), t is weaker than S ). This mandatory requirement simplifies the design of the transmission system between the processor and the memory, and improves the Data Reading speed. For example, a processor reads or writes 8 bytes of data at a time starting from an eight-fold address each time it reads/writes memory, if the software can ensure that data of the double type starts from an eight-fold address, it is only required to read or write data of the double type once.
Memory Operation. Otherwise, we may need two memory operations to complete this operation, because the data may be distributed across two 8-byte memory blocks that meet the alignment requirements. Some Processors may encounter errors when the data does not meet the requirements for alignment, But Intel's ia32 architecture processor can work correctly regardless of whether the data is aligned. However, Intel recommends that if you want to improve performance, all program data should be aligned as much as possible.

The ansi c standard does not stipulate that variables declared adjacent must be adjacent in memory. For program efficiency, memory alignment problems are flexibly handled by the compiler, which may cause some padding bytes between adjacent variables. For the basic data type (INT char), the memory space they occupy has a fixed value in a fixed hardware system. Therefore, we will only consider the memory allocation of struct members.

Alignment policy of Microsoft C compiler (cl.exe for 80 × 86) in win32platform:
1) The first address of the struct variable can be divisible by the size of its widest basic type member;
Note: When the compiler opens space for the struct, it first finds the widest basic data type in the struct, and then finds the location where the memory address can be divisible by the basic data type, the first address of the struct. Use the size of the widest basic data type as the alignment modulus described above. 
2) The offset (offset) of each member of the struct to the first address of the struct is an integer multiple of the member size. If necessary, the compiler will add the internal adding between the members );
Note: before opening a space for a member of the struct, the compiler first checks whether the offset of the first address of the pre-opening space to the first address of the struct is an integer multiple of the current member. If yes, It stores the member, on the contrary, a certain number of bytes are filled between the current member and the previous Member to meet the integer double requirement, that is, the first address of the pre-opened space is removed several bytes. 
3) the total size of the struct is an integer multiple of the size of the widest basic type of the struct. If necessary, the compiler will add the trailing padding after the last member ).
Note: The total size of the struct includes the padding byte. The last member must meet the preceding two conditions and the third condition. Otherwise, the last few bytes must be filled to meet the requirements.

According to the preceding rules, in windows, the size of sizeof (t) is 8 bytes using the VC compiler.

In the GNU gcc compiler, there are some differences in the principles followed. The alignment modulus is not determined based on the widest basic data type as described above. In GCC, the maximum alignment modulus is 4. That is to say, the alignment modulus can only be 1, 2, and 4, even if the structure has a double type. In addition, in the preceding three items, the offset value must be an integer multiple of the member size. If the member size is smaller than or equal to 4, the offset value is calculated according to the preceding rules. If the member size is greater than 4, the offset of each member of the struct to the first address of the struct can only be determined by an integer multiple of 4.
Take the following example:
Struct t
{
Char ch;
Double D;
};
In GCC, sizeof (t) should be 12 bytes.

If the struct contains bit-field, the guidelines in VC must be changed:
1) if the types of adjacent fields are the same, and the sum of the bit widths is smaller than the sizeof size of the type, the subsequent fields will be stored next to the previous field until they cannot be accommodated;
2) If the Field Types of adjacent bit fields are the same, but the sum of Bit Width is greater than the sizeof size of the type, the subsequent fields start from the new storage unit, its offset is an integer multiple of its type;
3) if the types of adjacent bitfield fields are different, the specific implementation of each compiler varies, vc6 adopts the non-compression mode (the fields of different bit domains are stored in different bit domain type bytes), and both Dev-C ++ and GCC adopt the compression mode;
Note: When the two fields are of different types, for example:
Struct n
{
Char C: 2;
Int I: 4;
};
The memory alignment criteria for the non-bit domain struct are still met. the offset of the I member to the first address of the struct should be an integer multiple of 4. Therefore, the C member must be filled with three bytes, then the space of four bytes is opened up as the int type, four of which are used to store I, so the space occupied by the above struct in VC is 8 bytes; for compilers that adopt compression, the memory alignment criteria of the non-bit domain structure are followed. The difference is that if the three words are filled with energy saving, the data is compressed to the padding byte, which cannot be accommodated. Therefore, the space occupied by the above struct N in GCC or Dev-C ++ should be 4 bytes.

4) do not compress fields that are interspersed with non-bit fields;
Note:
Struct
Typedef struct
{
Char C: 2;
Double I;
Int C2: 4;
} N3;
The space occupied by GCC is 16 bytes, and the space occupied by VC is 24 bytes.
5) the total size of the entire struct is an integer multiple of the size of the widest basic type.

PS:

The choice of alignment modulus can only be based on the basic data type. Therefore, for the nested struct In the struct, you can only consider the basic data type to be split. For the 2nd records in the alignment criterion, the entire struct is regarded as a member. The size of the member is determined based on the alignment criterion.

Class objects are stored in the memory in a similar way as struct, which is not described here. It should be noted that the size of the class object only includes the space occupied by non-static member variables of the class. If there is a virtual function, you can add another space occupied by the pointer.

For example

Union date
{
Double I;
Int K [5];
Char C;
};

Struct data
{
Char cat;
Union date cow;
Char dog;
} Too;

Printf ("% d \ n", sizeof (struct data), sizeof (max ));

Output: 40 24

This is because the alignment modulus of the struct is 8 (double) for the data sharing body. However, due to the characteristics of the sharing body, the allocation space should be allocated based on int K [5, then we should allocate 20 + 4 (Structure alignment) = 24 bytes.

In the data structure, the Union data cow should be used as the alignment modulus based on the basic data type (double) in the Union data, which is obviously 8. the entire struct Data Alignment mode should also be 8. therefore, data occupies 8 + 24 + 8 = 40 bytes.