aline alignment

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. In practice, the answer is actually 8 bytes tested in vc6. Well, I am always hurt, and I am a little nu

For ia32 commands, the alignment policy used by Linux is: the address of the 2-byte data type must be a multiple of 2, while the larger data types (INT, int *, float, and double) must be a multiple of 4. Stricter requirements for Microsoft Windows alignment -------- the address of any k-byte basic object must be a multiple of K, K = 2, 4, or 8. In fact, after extending to the X86-64 directive, it's all the

(Nonsense: I recently came into contact with the memory alignment concept. For memory alignment rules, I verified them one by one in VC and found that the compiler indeed alignment my data in memory. I know that memory alignment must have its advantages. What is it? I am looking for materials to say that memory

Byte Length, byte alignment, class, object Length struct st {short num;float math_grade;float Chinese_grade;float sum_grade;};int _tmain(int argc, _TCHAR* argv[]){cout ---------------------------- The result are :------------------------------------- To enable the CPU to efficiently and quickly access variables, the starting address of the variable should have some characteristics, that is, the so-called "

At the time of storage, in order to improve the efficiency, the offset is usually placed in the position of 2 m, and it is often rounded up and down to take the whole two requirements.Rounding downPalign_down (X,align) (X (-align))Why do you do this, because align or oneself, only the high point all become 1, and then the original number operation, at this time not more than 1 are cleared 0.PALIGN_UP (X,align) (-(x) (-align))The principle is also relatively easy to infer, the X into a negativ

Byte alignment detailsArticleSee http://blog.csdn.net/weixing979/archive/2006/05/12/725638.aspx Where a struct is a member of B structCalculate the length of a and the valid alignment value of A, and then perform the filling calculation. The following struct # Pragma pack (8)Struct S1{Short;Long B;}; Struct S2{Char C;S1 D;}; S1 is a member of S2. The S1 length is 8, and the valid

Supported but deprecated directives in 1:GCC#pragma pack (): Cancels memory-aligned access#pragma pack (n) (N=1/2/4/8): aligned by N-byte#pragma pack (2) struct mystruct1{int A;char b;short c;} struct mystruct2{int A;;Double B;short C;} #pragma pack ()This part of the content is 2-byte aligned.Analysis:(1) #pragma是用来指挥编译器, or set the alignment of the compiler. The default alignment for the compiler is 4, bu

Alignment issues for basic data types:Where variables are stored in memory generally requires natural alignment. The so-called natural alignment , is that the basic data type of the variable can not be stored simply in memory arbitrary position, but its starting address must meet can be divisible by their size. For example, 32-bit platforms, the address of int an

# Pragma pack (8) Struct S1 {Short;Long B;}; Struct S2 {Char C;S1 D;Long long E;}; # Pragma pack () Question1. sizeof (S2) =?2. How many bytes are left behind C in S2 and then D? 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, n

{Char C;Int I;}; What is sizeof (S1) equal? Clever, you start to think about it. Char occupies 1 byte and INT occupies 4 byte, so the total value should be 5. Is that true? Have you tried it on your machine? Maybe you are right, but you are probably wrong! In vc6, the default value is 8. Why? Why is my injury always me? Don't be frustrated. Let's take a closer look at the definition of sizeof. The result of sizeof is equal to the number of memory bytes occupied by objects or types. Well, let's t

http://blog.csdn.net/xing_hao/article/details/6678048One. Memory alignmentMany computer systems restrict the location of basic types of data in memory, and they require that the value of the first address of the data be a multiple of a number k (usually 4 or 8), which is called memory alignment, and this k is called the Zimo number of the data type (alignment modulus). When the ratio of one type S pair of Z

What is byte alignment and why is it aligned? The memory space in modern computers is divided by byte, theoretically, it seems that access to any type of variable can begin at any address, but the reality is that access to a particular type of variable is often at a specific memory address, which requires that various types of data be arranged in space according to certain rules, Instead of sequencing one after another, that's the

This problem is indeed quite basic. In order to give yourself a summary, this article is also a note record and strives to be the most streamlined. Structure alignment keywords: Alignment coefficient (also called alignment modulus): This coefficient is changed using the pre-compiled command # pragma pack (N), n =, 16. "Ali