Visual c++ pragma包(8)应该如何工作?
我不熟悉结构对齐和包装。我以为我理解了,但我发现了一些我没有预料到的结果(见下文) 我对结构对齐的理解是:Visual c++ pragma包(8)应该如何工作?,visual-c++,struct,alignment,pragma-pack,Visual C++,Struct,Alignment,Pragma Pack,我不熟悉结构对齐和包装。我以为我理解了,但我发现了一些我没有预料到的结果(见下文) 我对结构对齐的理解是: 类型通常在大小为其倍数的内存地址上对齐 根据需要添加衬垫,以便于正确对齐 结构的末端必须填充到最大元素的倍数(以便于阵列访问) #pragma pack指令基本上允许覆盖基于类型大小的对齐的一般约定: #pragma pack(push, 8) struct SPack8 { // Assume short int is 2 bytes, double is 8 bytes, and
#pragma pack
指令基本上允许覆盖基于类型大小的对齐的一般约定:
#pragma pack(push, 8)
struct SPack8
{
// Assume short int is 2 bytes, double is 8 bytes, and int is 4 bytes
short int a;
double b;
short int c;
int d;
};
#pragma pack(pop)
Pseudo struct layout: What I expected:
// note: PADDING IS BRACKETED
0, 1, [2, 3, 4, 5, 6, 7] // a occupies address 0, 1
8, 9, 10, 11, 12, 13, 14, 15, // b occupies 8-15 inclusive
16, 17, [18, 19, 20, 21, 22, 23] // c occupies 16-17 inclusive
24, 25, 26, 27 // d occupies 24-27 inclusive
// Thus far, SPack8 is 28 bytes, but the structure must be a multiple of
// sizeof(double) so we need to add padding to make it 32 bytes
[28, 29, 30, 31]
令我惊讶的是,在VS 2015 x86上,sizeof(SPack8)=24。d似乎没有在8字节地址上对齐:
offsetof(SPack, a) // 0, as expected
offsetof(SPack, b) // 8, as expected
offsetof(Spack, c) // 16, as expected
offsetof(SPack, d) // 20..what??
有人能解释一下发生了什么/我误解了什么吗
谢谢大家! 您的误解是,
#pragma pack
允许您扩展结构,但它不能<代码>打包允许您在需要时更紧密地打包结构。#pragma包(push,8)
告诉编译器,它最多只能在8字节边界上对齐,但不能超过
例如:
#pragma pack(push, 2)
struct X {
char a; // 1 byte
// 1 byte padding
int b; // 4 bytes, note though that it's aligned on 2 bytes, not 4.
char c, d, e; // 3 bytes
//1 byte padding
}; // == 10 bytes, the whole struct is also aligned on 2 bytes, not 4
#pragma pack(pop)
// The same struct without the pragma pack:
struct Y {
char a; // 1 byte
// 3 bytes padding
int b; // 4 bytes
char c, d, e; // 3 bytes
// 1 byte padding
};
这就是pack
所做的,编译器通常使用较少的填充。在您的示例中,您尝试在8字节边界上对齐int
,但由于您允许编译器最多在8字节上对齐,因此编译器希望使用的4字节对齐是可以的。您的整个结构(大小为24)的大小也是8的倍数(您的最大成员),因此不需要填充到32
可以强制对齐结构
__declspec(align(32)) struct Z {
char a;
int b;
char c, d, e;
};
struct SPack8
{
// Assume short int is 2 bytes, double is 8 bytes, and int is 4 bytes
short int a;
double b;
short int c;
__declspec(align(8)) int d;
};
甚至是你的结构的一个成员
__declspec(align(32)) struct Z {
char a;
int b;
char c, d, e;
};
struct SPack8
{
// Assume short int is 2 bytes, double is 8 bytes, and int is 4 bytes
short int a;
double b;
short int c;
__declspec(align(8)) int d;
};
在特定的边界上,但我不认为有理由强制将4字节类型与8字节对齐