The byte alignment principle of the "go" c-language structure

Original link: http://blog.csdn.net/shenbin1430/article/details/4292463

Why are you aligning?

The memory space in modern computers is divided by Byte, in theory it seems that access to any type of variable can start from any address, but the reality is that when accessing a particular type of variable, it is often accessed at a specific memory address, which requires all types of data to be spatially arranged according to certain rules, Instead of sequentially one by one emissions, that's the alignment.

The effect and reason of alignment: the processing of storage space varies greatly with each hardware platform. Some platforms can only access certain types of data from certain specific addresses. For example, some architectures have an error when accessing a variable that is not aligned, so programming must ensure byte alignment in this architecture. Other platforms may not have this, but the most common is the loss of access efficiency if the data storage is not aligned according to their platform requirements. For example, some platforms read each time from the even address, if an int (assuming 32-bit system) if the location of the place where the even address begins, then a read cycle can be read out of the 32bit, and if it is stored at the beginning of the odd address, it takes 2 read cycles, The 32bit data can be obtained by piecing together the high and low bytes of the two read-out results. Obviously, the reading efficiency is much lower.

Two. The effect of byte alignment on the program:

Let's take a look at some examples (32bit,x86 environment, GCC compiler):

The structure is defined as follows:

struct A

{

int A;

Char b;

Short C;

};

struct B

{

Char b;

int A;

Short C;

};

The lengths of the various data types on the 32-bit machine are now known as:

Char:1 (Signed and unsigned)

Short:2 (Signed and unsigned)

Int:4 (Signed and unsigned)

Long:4 (Signed and unsigned)

Float:4 Double:8

So what's the size of the top two structures?

The result is:

sizeof (Strcut A) value is 8

sizeof (struct B) has a value of 12.

struct A contains 4-byte-length int one, 1-byte-length char one and 2-byte length of short data one, B is the same;

The result above is because the compiler wants to align the data members spatially. The above is the result of the alignment according to the compiler's default settings, then we can not change the compiler's default alignment settings, of course. For example:

#pragma pack (2)/* Specify 2-byte alignment */

struct C

{

Char b;

int A;

Short C;

};

#pragma pack ()/* To cancel the specified alignment, restore the default alignment */

The sizeof (struct C) value is 8.

Modify the alignment value to 1:

#pragma pack (1)/* Specify 1-byte alignment */

struct D

{

Char b;

int A;

Short C;

};

#pragma pack ()/* To cancel the specified alignment, restore the default alignment */

The sizeof (struct D) value is 7.

Later we explain the role of #pragma pack ().

Three. What are the principles for the compiler to align?

Let's start by looking at four important basic concepts:

1. The alignment value of the data type itself:

For char type data, its own alignment value is 1, for the short type is 2, for the int,float,double type, its own alignment value is 4, the unit byte.

2. The self-aligning value of a struct or class: The value that is the largest of its members in its own alignment value.

3. Specify the alignment value: The specified alignment value when #pragma pack (value) is values.

4. Valid alignment values for data members, structs, and classes: The value that is small for its own alignment value and for the specified alignment value.

With these values, we can easily discuss the members of a specific data structure and its own alignment. The valid alignment value n is the value that is ultimately used to determine how the data is stored, most importantly. A valid alignment of n is the "alignment on n", which means that the data "holds the starting address%n=0". Data variables in the structure are emitted in a defined order of precedence. The starting address of the first data variable is the starting address of the structure. The member variables of the struct should be aligned with the emission, and the structure itself should be rounded according to its own valid alignment value (that is, the total length of the struct member variable is required to be an integral multiple of the effective alignment value of the struct, as the following example understands). This makes it impossible to understand the values of several examples above.

Example Analysis:

analysis example B;

struct B

{

Char b;

int A;

Short C;

};

Suppose B starts discharging from the address space 0x0000. In this example, the specified alignment value is not defined, and in the author's environment, the value defaults to 4. The first member variable B has its own alignment value of 1, which is smaller than the specified or default alignment value of 4, so its valid alignment value is 1, so its storage address 0x0000 conforms to 0x0000%1=0. The second member variable A has its own alignment value of 4, so the valid alignment value is also 4. So it can only be stored in the starting address of 0x0004 to 0x0007 four consecutive byte space, review 0x0004%4=0, and immediately close to the first variable. The third variable, C, has its own alignment value of 2, so the valid alignment value is also 2, which can be stored in the two byte space of 0x0008 to 0x0009, conforming to 0x0008%2=0. So everything from 0x0000 to 0x0009 is stored in B content. Then look at the data structure B's own alignment value for its variable maximum alignment value (here is B) so is 4, so the structure of the effective alignment value is also 4. 0x0009 to 0x0000=10 bytes, (10+2)%4=0 according to the requirements of the structural rounding. So 0x0000a to 0x000b is also occupied by struct B. So B has a total of 12 bytes from 0x0000 to 0x000b, sizeof (struct B) = 12; In fact, if this is the case it has aligned the satisfied byte, because its starting address is 0, so it must be aligned, the reason is to add 2 bytes later, Because the compiler is in order to achieve the access efficiency of the array of structures, imagine if we define an array of struct B, then the first struct starts with a 0 no problem, but the second structure? As defined by the array, all elements in the array are next to each other, and if we do not add the size of the structure to the integer multiples of 4, Then the starting address of the next structure will be 0x0000a, which obviously does not satisfy the structure's address alignment, so we want to add the structure to an integer multiple of the effective alignment size. In fact, for char type data, its own alignment value is 1, for the short type is 2, for Int,float, A double type with its own alignment value of 4, these existing types of self-aligning values are also based on arrays, only because the lengths of these types are known, so their own alignment values are known.

Similarly, analyze the above example C:

#pragma pack (2)/* Specify 2-byte alignment */

struct C

{

Char b;

int A;

Short C;

};

#pragma pack ()/* To cancel the specified alignment, restore the default alignment */

The first variable B has its own alignment value of 1, the specified alignment value is 2, so its valid alignment value is 1, assuming that C starts with 0x0000, then B is stored in 0x0000, conforms to 0x0000%1= 0, the second variable has its own alignment value of 4, and the alignment value is 2, so the valid alignment value is 2. So the order is stored in 0x0002, 0x0003, 0x0004, 0x0005 four consecutive bytes, in accordance with 0x0002%2=0. The third variable, C, has its own alignment value of 2, so the valid alignment value is 2, which is stored sequentially

In 0x0006, 0x0007, in accordance with 0x0006%2=0. So from 0x0000 to 0x00007 a total of eight bytes is stored in the C variable. and C has its own alignment value of 4, so the valid alignment value for C is 2. And 8%2=0,c only takes up eight bytes of 0x0000 to 0x0007. So sizeof (struct C) =8.

Four. How do I modify the compiler's default alignment values?

1. In the VC IDE, you can modify this: [project]| [settings],c/c++ tab category of the Code generation option of the struct Member alignment modified, the default is 8 bytes.

2. When encoding, you can change this dynamically: #pragma pack. Note: It is pragma instead of progma.

Five. What do we consider in programming for byte alignment?

If you want to consider saving space when programming, then we only need to assume that the first address of the structure is 0, then the various variables are arranged according to the above principles, the basic principle is to make the variables in the structure from small to large declaration of the type size, Minimize the amount of space in the middle. There is another way to make space for the efficiency of time, we show the space to be filled to align, such as: There is a space-time approach is to explicitly insert reserved members:

struct a{

Char A;

Char reserved[3];//use space to change time

int b;

}

The reserved member has no meaning to our program, it just fills the space to achieve byte alignment, and of course even without this member the compiler will automatically fill us with the alignment, and we add it just to play an explicit reminder.

Six. Possible pitfalls of byte alignment:

Many of the pitfalls of alignment in code are implicit. For example, when forcing type conversions. For example:

unsigned int i = 0x12345678;

unsigned char *p=null;

unsigned short *p1=null;

p=&i;

*p=0x00;

p1= (unsigned short *) (p+1);

*p1=0x0000;

The last two lines of code, from the odd boundary to access the Unsignedshort variable, clearly do not conform to the rules of alignment.

On x86, similar operations can only affect efficiency, but on MIPS or SPARC, it can be an error because they require byte alignment.

Seven. How to find problems with byte alignment:

If an alignment or assignment problem occurs, first view

1. Compiler's big little side settings

2. See if the system itself supports non-aligned access

3. If the support depends on whether the alignment is set or not, if there is no access, you need to add some special decorations to flag its special access operation.

Alignment processing under Arm

From Dui0067d_ads1_2_complib

3.13 Type Qulifiers

Partial excerpt from ARM compiler document alignment Section

Use of alignment:

1.__align (num)

This is used to modify the byte bounds of the highest-level object. When using LDRD or STRD in the assembly

Use this command __align (8) to make a modifier limit. To ensure that the data Objects are aligned accordingly.

The maximum command for this decorated object is a 8-byte limit, which allows 2-byte objects to be 4 bytes

aligned, but cannot align 4 bytes of object 2 bytes.

__align is a storage class modification, and he only modifies the top type objects that cannot be used for structs or function objects.

2.__packed

__packed is a one-byte alignment

1. Packed objects cannot be aligned

2. Non-aligned access to read and write access to all objects

3.float and the structure containing float and unused __packed objects will not be byte-aligned

4.__packed has no effect on local shaping variables

5. Forcing conversion from unpacked object to packed object is undefined, the shaping pointer can be legally fixed

Righteousness is packed.

__packed int* p; __packed int is meaningless

6. Alignment or non-aligned read-write access poses a problem

__packed struct Struct_test

{

Char A;

int b;

char c;

}  ; Define the following structure at this point, the start address of B must be misaligned.

There may be a problem accessing B in the stack because the data on the stack is definitely aligned access [from CL]

Define the following variables as global static not on the stack

Static char* p;

static struct struct_test A;

void Main ()

{

__packed int* Q; This is defined as __packed to decorate the access below the current Q point to a non-aligned data address, which can be

p = (char*) &a;

Q = (int*) (p+1);

*q = 0x87654321;

/*

The assembly instructions that get the assignment are clear.

LDR r5,0x20001590; = #0x12345678

[0xe1a00005] mov r0,r5

[0xeb0000b0] BL __rt_uwrite4//Call a write 4byte operation function here

[0xe5c10000] Strb r0,[r1, #0]//function to perform 4 STRB operations and then return to ensure the correct access to the data

[0xe1a02420] mov r2,r0,lsr #8

[0xe5c12001] Strb r2,[r1, #1]

[0xe1a02820] mov r2,r0,lsr #16

[0xe5c12002] Strb r2,[r1, #2]

[0XE1A02C20] mov r2,r0,lsr #24

[0xe5c12003] Strb r2,[r1, #3]

[0xe1a0f00e] mov pc,r14

*/

/*

If Q does not have a __packed modifier then the assembler command is such that direct access to the singular address fails

[0xe59f2018] Ldr r2,0x20001594; = #0x87654321

[0xe5812000] str r2,[r1, #0]

*/

This makes it clear how non-aligned access produces errors.

And how to eliminate non-aligned access poses problems

You can also see non-aligned Access and aligned access instruction differences leading to efficiency issues

}

The determination of the size of the structure by sizeof

typedef struct

{

int A;

Char b;

}a_t;

typedef struct

{

int A;

Char b;

char c;

}b_t;

typedef struct

{

Char A;

int b;

char c;

}c_t;

void Main ()

{

char*a=0;

Cout<<sizeof (a) <<ENDL;//4

Cout<<sizeof (*a) <<endl;//1--This can understand

Cout<<sizeof (a_t) <<ENDL;//8

Cout<<sizeof (b_t) <<ENDL;//8

Cout<<sizeof (c_t) <<ENDL;//12

}

Why is this the result?

2. Syntax:

sizeof has three grammatical forms, as follows:

1) sizeof (object); sizeof (object);

2) sizeof (TYPE_NAME); sizeof (type);

3) sizeof object; sizeof object;

5. sizeof for pointer variables

Since it is to store the address, it certainly equals the width of the internal address bus of the computer. So in a 32-bit computer, a

The return value of a pointer variable must be 4 (in bytes), which can be expected in a future 64-bit system

The sizeof result in pointer variable is 8.

char* PC = "ABC";

int* Pi;

string* PS;

char** PPC = &pc;

void (*PF) ();//function pointer

sizeof (PC); Result is 4

sizeof (PI); Result is 4

sizeof (PS); Result is 4

sizeof (PPC); Result is 4

sizeof (PF);//result is 4

The sizeof value of the pointer variable has nothing to do with the object that the pointer refers to, because all the pointer variables occupy memory

are equal in size, so the MFC message processing function uses two parameters wparam, LPARAM can pass a variety of complex message knot

Use a pointer to the struct body.

6. The sizeof array

The sizeof value of an array is equal to the number of bytes of memory consumed by the array, such as:

Char a1[] = "ABC";

int a2[3];

sizeof (A1); The result is 4, and there is a null terminator at the end of the string

sizeof (A2); Result is 3*4=12 (dependent on int)

Some friends started with sizeof as the number of array elements, and now you should know it's not right, that

How should we find the number of array elements? Easy, usually has the following two kinds of wording:

int c1 = sizeof (A1)/sizeof (char); Total length/length of individual elements

int c2 = sizeof (A1)/sizeof (a1[0]); Total length/length of first element

Write here, ask, the following C3,C4 value should be how much?

void Foo3 (char a3[3])

{

int c3 = sizeof (A3); C3 = =

}

void Foo4 (char a4[])

{

int c4 = sizeof (A4); C4 = =

}

Maybe when you try to answer C4 's value, you realize that C3 answered wrong, yes, c3!=3. Here the function parameter A3 is no longer

Array type, instead of transforming into pointers, equivalent to char* A3, why? Think it over and it's easy to understand that we call

When the function foo1, will the program allocate an array of size 3 on the stack? No! The array is "transmitted", and the caller

Simply pass the address of the argument to the past, so A3 is naturally the pointer type (char*), and the C3 value is 4.

7. sizeof of the structure

This is a question that beginners ask most, so it is necessary to pay more and more ink. Let's look at a structure first:

struct S1

{

char c;

int i;

};

Q. How much does sizeof (S1) equal? Smart you start thinking, Char takes 1 bytes, int is 4 bytes, then add

It's supposed to be 5. Is that right? Have you tried it on your machine? Maybe you're right, but you're probably wrong! V

The result of the default setting in C6 is 8.

Why? Why is it always me that hurts?

Please don't be upset, let's have a good look at the definition of sizeof the result of--sizeof is equal to the object or type.

Memory bytes, okay, let's take a look at the memory allocations for S1:

S1 S1 = {A, 0xFFFFFFFF};

After defining the above variables, add breakpoints, run the program, observe the memory where the S1 is located, what do you find?

Take my VC6.0 for example, S1 's address is 0x0012ff78, and its data is as follows:

0012ff78:61 cc CC FF FF FF FF

What did you find out? How is the 3-byte cc in the middle? Check out the instructions on MSDN:

When applied to a structure type or variable, sizeof returns the actual size,

Which may include padding bytes inserted for alignment.

So this is the legendary byte alignment! An important topic came up.

Why do I need byte alignment? The principle of computer composition teaches us that this helps to speed up the number of computers, otherwise

You have to spend more time on the instruction cycle. To do this, the compiler handles the structure by default (in fact the data in other places is changed

), the base data type (short, etc.) with a width of 2 is located at an address divisible by 2, allowing the width

The basic data type of 4 (int, etc.) is located on an address that is divisible by 4, and so on. In this way, the middle of the two-digit

Padding bytes may need to be added, so the sizeof value of the entire struct is increased.

Let's swap the position of char and int in S1:

struct S2

{

int i;

char c;

};

Look at the results of sizeof (S2), how much is 8? Look at the memory again, the original member C still has 3 filler words behind

section, which is why AH? Don't worry, summarize the rules below.

Byte alignment details are related to compiler implementations, but in general, three guidelines are met:

1) The first address of a struct variable can be divisible by the size of its widest base type member;

2) the offsets (offset) of each member of the struct relative to the first address of the struct are an integer multiple of the member size, if any

Requires the compiler to add padding bytes (internal adding) between members;

3) The total size of the struct is an integer multiple of the size of the structure's widest base type member, and if necessary, the compiler will

Member and then add padding bytes (trailing padding).

For the above guidelines, there are a few things to note:

1) The first is not to say that the address of a struct member is an integer multiple of its size, how is it said that the offset amount? Because with the first

1 points exist, so we can just consider the offset of the member, so it is simple to think. Think about why.

The offset of a member of a struct relative to the first address of the struct can be obtained by using the macro Offsetof (), which is also

Stddef.h is defined in the following:

#define OFFSETOF (S,m) (size_t) & (((S *) 0)->m)

For example, to get the offset of C in S2, the method is

size_t pos = Offsetof (S2, c);//POS equals 4

2) The base type refers to the previously mentioned built-in data types such as char, short, int, float, double,

The "data width" here refers to the size of its sizeof. Because a member of a struct can be a composite type,

As another struct, so when looking for the widest base type member, you should include the child members of the compound type member,

Instead of seeing composite members as a whole. However, when determining the offset position of a composite type member, the composite type

As a whole view.

Here is a bit of a mouthful, thinking is also a bit scratching their heads, or let us take a look at the example (the exact value is still VC

6 For example, no longer explained later):

struct S3

{

Char C1;

S1 s;

Char C2

};

The type of the widest simple member of the S1 is int,s3 to "break" the S1 when considering the widest simple type member, so

The widest simple type of S3 is int, so that a variable defined by S3 needs to be divisible by 4 for the first address of the storage space, the entire

The value of a sizeof (S3) should also be divisible by 4.

Is the offset of the C1 0,s? At this point S is a whole, and as a struct variable also satisfies the previous three criteria

, so its size is 8, the offset is between 4,C1 and s needs 3 padding bytes, and C2 and s no need,

So the C2 offset is 12, the size of the C2 is 13,13 can not be divisible by 4, so the end of the 3 fill

byte-filling. Finally, the value of sizeof (S3) is 16.

From the above narrative, we can get a formula:

The size of the struct is equal to the last member's offset plus its size plus the number of padding bytes at the end, namely:

sizeof (struct) = Offsetof (last item) + sizeof (last item) + sizeof (trail

ing padding)

Here, friends should have a whole new understanding of the structure of sizeof, but do not be happy too early, there is a

The important parameters affecting sizeof have not been mentioned, which is the compiler's pack instructions. It is used to adjust the alignment of the structure body

, different compiler names and usages are slightly different, implemented by #pragma pack in VC6, and can be modified directly

/ZP the compile switch. The basic usage of #pragma pack is: #pragma pack (n), n is a byte-aligned number, and its value

is 1, 2, 4, 8, 16, the default is 8, if this value is smaller than the sizeof value of the struct member, then the member's offset

The amount should be based on this value, that is, the structure member of the offset should take the minimum value, the formula is as follows:

Offsetof (item) = MIN (n, sizeof (item))

Look again at the example:

#pragma pack (push)//Save current pack settings to stack

#pragma pack (2)//must be used before structure definition

struct S1

{

char c;

int i;

};

struct S3

{

Char C1;

S1 s;

Char C2

};

#pragma pack (POP)//Restore Previous pack settings

When sizeof (S1) is calculated, the value of min (2, sizeof (i)) is 2, so the offset of I is 2, plus sizeof (i) equals

6, can be divisible by 2, so the size of the whole S1 is 6.

Similarly, for sizeof (S3), the offset of S to 2,C2 is 8, plus sizeof (C2) equals 9, cannot be

2 is divisible, adding a padding byte, so sizeof (S3) equals 10.

Now, friends can easily breathe out,

It is also important to note that the size of the "empty struct" (without data members) is not 0, but 1. Imagine a "not accounted for

How can the variable of space "be taken address, two different" empty struct "variables be distinguished? So, "

An empty struct "variable is also stored so that the compiler can allocate only one byte of space for the placeholder.

As follows:

struct S5 {};

sizeof (S5); Result is 1

8. sizeof with a bit domain structure

As mentioned earlier, bit-domain members cannot be individually taken with the sizeof value, and what we're going to discuss here is the struct with the bit field.

sizeof, only to consider its particularity and specifically listed it.

C99 specifies that int, unsigned int, and bool can be used as bit field types, but almost all compilers extend this to allow

Other types of type exist.

The primary purpose of using bit fields is to compress the storage, with the following approximate rules:

1) If the adjacent bit field field is of the same type and its bit width is less than the type sizeof size, the following field will

Adjacent to the previous field store until it cannot be accommodated;

2) If the adjacent bit field field has the same type, but its bit width is greater than the type sizeof size, then the following field will

Starting from the new storage unit, its offset is an integer multiple of its type size;

3) If there are different types of adjacent bit field fields, the specific implementation of each compiler will differ, VC6 take the non-compression method

, dev-c++ adopt compression method;

4) If the bit field fields are interspersed with non-bit field fields, no compression is performed;

5) The total size of the entire struct is an integer multiple of the size of the widest base type member.

Let's take a look at the example.

Example 1:

struct BF1

{

Char F1:3;

Char F2:4;

Char F3:5;

};

Its memory layout is:

|_f1__|__f2__|_|____f3___|____|

|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|

0 3 7) 8 1316

The bit field type is char, the 1th byte can only hold the lower F1 and F2, so F2 is compressed into the 1th byte, and F3 can only

Start with the next byte. Thus the result of sizeof (BF1) is 2.

Example 2:

struct BF2

{

Char F1:3;

Short F2:4;

Char F3:5;

};

Due to the different types of neighboring bit fields, the sizeof is 6 in VC6 and 2 in dev-c++.

Example 3:

struct BF3

{

Char F1:3;

char F2;

Char F3:5;

};

Non-bit field fields are interspersed in which no compression is generated, and the size obtained in VC6 and Dev-c++ is 3.

9. SizeOf of the Consortium

Structure in the memory organization is sequential, the union is overlapping, each member shares a piece of memory, so the entire

The Fit sizeof is the maximum value of each member sizeof. A member of a struct can also be a composite type, here,

Composite type members are considered as a whole.

So, in the example below, the sizeof value of U equals sizeof (s).

Union U

{

int i;

char c;

S1 s;

};

The byte alignment principle of the "go" c-language structure