考虑:
struct mystruct_A
{
char a;
int b;
char c;
} x;
struct mystruct_B
{
int b;
char a;
} y;
结构尺寸分别为12和8。
这些结构是填充的还是包装的?
什么时候进行填充或包装?
当前回答
(上面的答案解释的很清楚，但是对于padding的大小似乎不是很清楚，所以，我将根据我从The Lost Art of Structure Packing学到的补充一个答案，它已经发展到不局限于C，也适用于Go, Rust。)
内存对齐(用于struct)
规则:
Before each individual member, there will be padding so that to make it start at an address that is divisible by its alignment requirement. E.g., on many systems, an int should start at an address divisible by 4 and a short by 2. char and char[] are special, could be any memory address, so they don't need padding before them. For struct, other than the alignment need for each individual member, the size of whole struct itself will be aligned to a size divisible by strictest alignment requirement of any of its members, by padding at end. E.g., on many systems, if struct's largest member is int then by divisible by 4, if short then by 2.
会员顺序:
成员的顺序可能会影响结构的实际大小，所以要记住这一点。 例如，下面示例中的stu_c和stu_d具有相同的成员，但顺序不同，并导致两个结构体的大小不同。
内存中的地址(用于struct)
空的空间:
两个结构体之间的空白空间可以被适合的非结构变量使用。 例如，在下面的test_struct_address()中，变量x位于相邻的结构g和h之间。 无论是否声明x, h的地址都不会改变，x只是重用了g浪费的空间。 y也是一样。
例子
(适用于64位系统)
memory_align.c:
/**
* Memory align & padding - for struct.
* compile: gcc memory_align.c
* execute: ./a.out
*/
#include <stdio.h>
// size is 8, 4 + 1, then round to multiple of 4 (int's size),
struct stu_a {
int i;
char c;
};
// size is 16, 8 + 1, then round to multiple of 8 (long's size),
struct stu_b {
long l;
char c;
};
// size is 24, l need padding by 4 before it, then round to multiple of 8 (long's size),
struct stu_c {
int i;
long l;
char c;
};
// size is 16, 8 + 4 + 1, then round to multiple of 8 (long's size),
struct stu_d {
long l;
int i;
char c;
};
// size is 16, 8 + 4 + 1, then round to multiple of 8 (double's size),
struct stu_e {
double d;
int i;
char c;
};
// size is 24, d need align to 8, then round to multiple of 8 (double's size),
struct stu_f {
int i;
double d;
char c;
};
// size is 4,
struct stu_g {
int i;
};
// size is 8,
struct stu_h {
long l;
};
// test - padding within a single struct,
int test_struct_padding() {
printf("%s: %ld\n", "stu_a", sizeof(struct stu_a));
printf("%s: %ld\n", "stu_b", sizeof(struct stu_b));
printf("%s: %ld\n", "stu_c", sizeof(struct stu_c));
printf("%s: %ld\n", "stu_d", sizeof(struct stu_d));
printf("%s: %ld\n", "stu_e", sizeof(struct stu_e));
printf("%s: %ld\n", "stu_f", sizeof(struct stu_f));
printf("%s: %ld\n", "stu_g", sizeof(struct stu_g));
printf("%s: %ld\n", "stu_h", sizeof(struct stu_h));
return 0;
}
// test - address of struct,
int test_struct_address() {
printf("%s: %ld\n", "stu_g", sizeof(struct stu_g));
printf("%s: %ld\n", "stu_h", sizeof(struct stu_h));
printf("%s: %ld\n", "stu_f", sizeof(struct stu_f));
struct stu_g g;
struct stu_h h;
struct stu_f f1;
struct stu_f f2;
int x = 1;
long y = 1;
printf("address of %s: %p\n", "g", &g);
printf("address of %s: %p\n", "h", &h);
printf("address of %s: %p\n", "f1", &f1);
printf("address of %s: %p\n", "f2", &f2);
printf("address of %s: %p\n", "x", &x);
printf("address of %s: %p\n", "y", &y);
// g is only 4 bytes itself, but distance to next struct is 16 bytes(on 64 bit system) or 8 bytes(on 32 bit system),
printf("space between %s and %s: %ld\n", "g", "h", (long)(&h) - (long)(&g));
// h is only 8 bytes itself, but distance to next struct is 16 bytes(on 64 bit system) or 8 bytes(on 32 bit system),
printf("space between %s and %s: %ld\n", "h", "f1", (long)(&f1) - (long)(&h));
// f1 is only 24 bytes itself, but distance to next struct is 32 bytes(on 64 bit system) or 24 bytes(on 32 bit system),
printf("space between %s and %s: %ld\n", "f1", "f2", (long)(&f2) - (long)(&f1));
// x is not a struct, and it reuse those empty space between struts, which exists due to padding, e.g between g & h,
printf("space between %s and %s: %ld\n", "x", "f2", (long)(&x) - (long)(&f2));
printf("space between %s and %s: %ld\n", "g", "x", (long)(&x) - (long)(&g));
// y is not a struct, and it reuse those empty space between struts, which exists due to padding, e.g between h & f1,
printf("space between %s and %s: %ld\n", "x", "y", (long)(&y) - (long)(&x));
printf("space between %s and %s: %ld\n", "h", "y", (long)(&y) - (long)(&h));
return 0;
}
int main(int argc, char * argv[]) {
test_struct_padding();
// test_struct_address();
return 0;
}
执行结果- test_struct_padding():
stu_a: 8
stu_b: 16
stu_c: 24
stu_d: 16
stu_e: 16
stu_f: 24
stu_g: 4
stu_h: 8
执行结果- test_struct_address():
stu_g: 4
stu_h: 8
stu_f: 24
address of g: 0x7fffd63a95d0 // struct variable - address dividable by 16,
address of h: 0x7fffd63a95e0 // struct variable - address dividable by 16,
address of f1: 0x7fffd63a95f0 // struct variable - address dividable by 16,
address of f2: 0x7fffd63a9610 // struct variable - address dividable by 16,
address of x: 0x7fffd63a95dc // non-struct variable - resides within the empty space between struct variable g & h.
address of y: 0x7fffd63a95e8 // non-struct variable - resides within the empty space between struct variable h & f1.
space between g and h: 16
space between h and f1: 16
space between f1 and f2: 32
space between x and f2: -52
space between g and x: 12
space between x and y: 12
space between h and y: 8
因此，每个变量的地址起始为g:d0 x:dc h:e0 y:e8
其他回答
(上面的答案解释的很清楚，但是对于padding的大小似乎不是很清楚，所以，我将根据我从The Lost Art of Structure Packing学到的补充一个答案，它已经发展到不局限于C，也适用于Go, Rust。)
内存对齐(用于struct)
规则:
Before each individual member, there will be padding so that to make it start at an address that is divisible by its alignment requirement. E.g., on many systems, an int should start at an address divisible by 4 and a short by 2. char and char[] are special, could be any memory address, so they don't need padding before them. For struct, other than the alignment need for each individual member, the size of whole struct itself will be aligned to a size divisible by strictest alignment requirement of any of its members, by padding at end. E.g., on many systems, if struct's largest member is int then by divisible by 4, if short then by 2.
会员顺序:
成员的顺序可能会影响结构的实际大小，所以要记住这一点。 例如，下面示例中的stu_c和stu_d具有相同的成员，但顺序不同，并导致两个结构体的大小不同。
内存中的地址(用于struct)
空的空间:
两个结构体之间的空白空间可以被适合的非结构变量使用。 例如，在下面的test_struct_address()中，变量x位于相邻的结构g和h之间。 无论是否声明x, h的地址都不会改变，x只是重用了g浪费的空间。 y也是一样。
例子
(适用于64位系统)
memory_align.c:
/**
* Memory align & padding - for struct.
* compile: gcc memory_align.c
* execute: ./a.out
*/
#include <stdio.h>
// size is 8, 4 + 1, then round to multiple of 4 (int's size),
struct stu_a {
int i;
char c;
};
// size is 16, 8 + 1, then round to multiple of 8 (long's size),
struct stu_b {
long l;
char c;
};
// size is 24, l need padding by 4 before it, then round to multiple of 8 (long's size),
struct stu_c {
int i;
long l;
char c;
};
// size is 16, 8 + 4 + 1, then round to multiple of 8 (long's size),
struct stu_d {
long l;
int i;
char c;
};
// size is 16, 8 + 4 + 1, then round to multiple of 8 (double's size),
struct stu_e {
double d;
int i;
char c;
};
// size is 24, d need align to 8, then round to multiple of 8 (double's size),
struct stu_f {
int i;
double d;
char c;
};
// size is 4,
struct stu_g {
int i;
};
// size is 8,
struct stu_h {
long l;
};
// test - padding within a single struct,
int test_struct_padding() {
printf("%s: %ld\n", "stu_a", sizeof(struct stu_a));
printf("%s: %ld\n", "stu_b", sizeof(struct stu_b));
printf("%s: %ld\n", "stu_c", sizeof(struct stu_c));
printf("%s: %ld\n", "stu_d", sizeof(struct stu_d));
printf("%s: %ld\n", "stu_e", sizeof(struct stu_e));
printf("%s: %ld\n", "stu_f", sizeof(struct stu_f));
printf("%s: %ld\n", "stu_g", sizeof(struct stu_g));
printf("%s: %ld\n", "stu_h", sizeof(struct stu_h));
return 0;
}
// test - address of struct,
int test_struct_address() {
printf("%s: %ld\n", "stu_g", sizeof(struct stu_g));
printf("%s: %ld\n", "stu_h", sizeof(struct stu_h));
printf("%s: %ld\n", "stu_f", sizeof(struct stu_f));
struct stu_g g;
struct stu_h h;
struct stu_f f1;
struct stu_f f2;
int x = 1;
long y = 1;
printf("address of %s: %p\n", "g", &g);
printf("address of %s: %p\n", "h", &h);
printf("address of %s: %p\n", "f1", &f1);
printf("address of %s: %p\n", "f2", &f2);
printf("address of %s: %p\n", "x", &x);
printf("address of %s: %p\n", "y", &y);
// g is only 4 bytes itself, but distance to next struct is 16 bytes(on 64 bit system) or 8 bytes(on 32 bit system),
printf("space between %s and %s: %ld\n", "g", "h", (long)(&h) - (long)(&g));
// h is only 8 bytes itself, but distance to next struct is 16 bytes(on 64 bit system) or 8 bytes(on 32 bit system),
printf("space between %s and %s: %ld\n", "h", "f1", (long)(&f1) - (long)(&h));
// f1 is only 24 bytes itself, but distance to next struct is 32 bytes(on 64 bit system) or 24 bytes(on 32 bit system),
printf("space between %s and %s: %ld\n", "f1", "f2", (long)(&f2) - (long)(&f1));
// x is not a struct, and it reuse those empty space between struts, which exists due to padding, e.g between g & h,
printf("space between %s and %s: %ld\n", "x", "f2", (long)(&x) - (long)(&f2));
printf("space between %s and %s: %ld\n", "g", "x", (long)(&x) - (long)(&g));
// y is not a struct, and it reuse those empty space between struts, which exists due to padding, e.g between h & f1,
printf("space between %s and %s: %ld\n", "x", "y", (long)(&y) - (long)(&x));
printf("space between %s and %s: %ld\n", "h", "y", (long)(&y) - (long)(&h));
return 0;
}
int main(int argc, char * argv[]) {
test_struct_padding();
// test_struct_address();
return 0;
}
执行结果- test_struct_padding():
stu_a: 8
stu_b: 16
stu_c: 24
stu_d: 16
stu_e: 16
stu_f: 24
stu_g: 4
stu_h: 8
执行结果- test_struct_address():
stu_g: 4
stu_h: 8
stu_f: 24
address of g: 0x7fffd63a95d0 // struct variable - address dividable by 16,
address of h: 0x7fffd63a95e0 // struct variable - address dividable by 16,
address of f1: 0x7fffd63a95f0 // struct variable - address dividable by 16,
address of f2: 0x7fffd63a9610 // struct variable - address dividable by 16,
address of x: 0x7fffd63a95dc // non-struct variable - resides within the empty space between struct variable g & h.
address of y: 0x7fffd63a95e8 // non-struct variable - resides within the empty space between struct variable h & f1.
space between g and h: 16
space between h and f1: 16
space between f1 and f2: 32
space between x and f2: -52
space between g and x: 12
space between x and y: 12
space between h and y: 8
因此，每个变量的地址起始为g:d0 x:dc h:e0 y:e8
这件事没有但是!想要掌握这门学科必须做到以下几点:
细读埃里克·s·雷蒙德所著的《丢失的结构包装艺术》 看一下Eric的代码示例 最后但并非最不重要的是，不要忘记下面关于填充的规则，即结构体的对齐方式与最大类型的对齐方式一致 要求。
填充规则:
结构体的每个成员都应该位于能被其大小整除的地址。 填充在元素之间或结构的末尾插入，以确保满足此规则。这样做是为了硬件更容易和更有效地访问总线。 结构体末尾的填充是根据结构体最大成员的大小决定的。
规则二: 考虑下面的结构，
如果我们要为这个结构体创建一个数组(包含2个结构体)， 结束时不需要填充:
因此，struct的大小= 8字节
假设我们要创建另一个结构体，如下所示:
如果我们要创建这个结构体的数组， 最后需要填充的字节数有两种可能。
A.如果我们在末尾添加3个字节，并将其对齐为int而不是Long:
B.如果我们在末尾添加7个字节并将其对齐为Long:
第二个数组的起始地址是8(i)的倍数。e 24)。 struct的大小= 24字节
因此，通过将结构体的下一个数组的起始地址对齐为最大成员(i。E如果我们要创建这个结构体的数组，第二个数组的第一个地址必须从一个地址开始，该地址必须是该结构体最大成员的倍数。这里是24(3 * 8))，我们可以计算出最后所需的填充字节数。
只有当你告诉编译器显式地对结构进行打包时，才会进行结构打包。你看到的是填充。您的32位系统正在填充每个字段以字对齐。如果您告诉编译器打包结构，它们将分别为6和5字节。但是不要这样做。它不可移植，使编译器生成的代码更慢(有时甚至有bug)。
这些结构是填充的还是包装的?
它们填充。
最初想到的唯一可能是，如果char和int的大小相同，那么char/int/char结构的最小大小将不允许填充，int/char结构也是如此。
然而，这将要求sizeof(int)和sizeof(char)都为4(以获得12和8的大小)。由于sizeof(char)始终为1的标准保证了整个理论的崩溃。
如果char和int的宽度相同，那么大小将是1和1，而不是4和4。因此，为了得到12的大小，在最终字段之后必须有填充。
什么时候进行填充或包装?
只要编译器实现需要。编译器可以在字段之间和最后一个字段之后(但不能在第一个字段之前)插入填充。
这样做通常是为了性能，因为某些类型在特定边界上对齐时性能更好。甚至有一些架构会在你试图访问未对齐的数据时拒绝运行(即崩溃)(是的，我在看你，ARM)。
您通常可以使用特定于实现的特性(如#pragma pack)来控制打包/填充(这实际上是同一领域的两个极端)。即使您不能在特定的实现中这样做，您也可以在编译时检查代码以确保它满足您的需求(使用标准C特性，而不是特定于实现的东西)。
例如:
// C11 or better ...
#include <assert.h>
struct strA { char a; int b; char c; } x;
struct strB { int b; char a; } y;
static_assert(sizeof(struct strA) == sizeof(char)*2 + sizeof(int), "No padding allowed");
static_assert(sizeof(struct strB) == sizeof(char) + sizeof(int), "No padding allowed");
如果这些结构中有任何填充，类似这样的东西将拒绝编译。
