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« Ответ #3 : 01-12-2003 06:24 » |
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ещё есть declspec(align(#bytes)). Появился в VC 7.0. Аргумент - кол-во байт, обязательно степень двойки от 1 до 8192 (pragma pack кончается на 8).
Удобен ещё и тем, что можно узнать выравнивание переменной или типа оператором __alignof(имя)
Цитата из МСДН 2003 (с полезным примером выравнивания по размеру кэши): ... The following examples show how __declspec(align(#)) affects the size and alignment of data structures. The examples assume the following definitions:
#define CACHE_LINE 32 #define CACHE_ALIGN __declspec(align(CACHE_LINE)) In the following example, the S1 structure is defined with __declspec(align(32)). All uses of S1, whether for a variable definition or other type declarations, ensure that this structure data is 32-byte aligned. sizeof(struct S1) returns 32, and S1 has 16 padding bytes following the 16 bytes required to hold the four integers. Each int member requires 4-byte alignment, but the alignment of the structure itself is declared to be 32, so the overall alignment is 32.
struct CACHE_ALIGN S1 // cache align all instances of S1{ int a, b, c, d; }; struct S1 s1; // s1 is 32-byte cache aligned In the following example, sizeof(struct S2) will return 16, which is exactly the sum of the member sizes, because that happens to be a multiple of the largest alignment requirement (a multiple of 8).
__declspec(align(8)) struct S2 { int a, b, c, d; }; In the following example, sizeof(struct S3) returns 64.
struct S3 { struct S1 s1; // S3 inherits cache alignment requirement // from S1 declaration int a; // a is now cache aligned because of s1 // 28 bytes of trailing padding }; In the following example, note that a has only the alignment of natural type, in this case, 4 bytes. However, S1 must be 32-byte aligned. Twenty-eight bytes of padding follow a, so that s1 starts at offset 32. S4 then inherits the alignment requirement of S1, because it is the largest alignment requirement in the structure. sizeof(struct S4) returns 64.
struct S4 { int a; // 28 bytes padding struct S1 s1; // S4 inherits cache alignment requirement of S1 }; The following three variable declarations also use __declspec(align(#)). In each case, the variable must be 32-byte aligned. In the case of the array, the base address of the array, not each array member, is 32-byte aligned. The sizeof value for each array member is not affected by the use of __declspec(align(#)).
CACHE_ALIGN int i; CACHE_ALIGN int array[128]; CACHE_ALIGN struct s2 s; To align each individual member of an array, code such as the following should be used:
typedef CACHE_ALIGN struct { int a; } S5; S5 array[10]; In the following example, note that aligning the structure itself and aligning the first element are identical:
CACHE_ALIGN struct S6 { int a; int b; };
struct S7 { CACHE_ALIGN int a; int b; }; S6 and S7 have identical alignment, allocation, and size characteristics.
Defining New Types with __declspec(align(#)) You can define a type with an alignment characteristic.
For example, you can define a struct with an alignment value as follows:
struct aType {int a; int b;}; typedef __declspec(align(32)) struct aType bType; Now, aType and bType are the same size (8 bytes) but variables of type bType will be 32-byte aligned.
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