__attribute__ ((__packed__))对嵌套结构有什么影响?例如:
// C version
struct __attribute__ ((__packed__))
{
struct
{
char c;
int i;
} bar;
char c;
int i;
} foo;
// C++ version
struct __attribute__ ((__packed__)) Foo
{
struct Bar
{
char c;
int i;
} bar;
char c;
int i;
} foo;我知道foo会很拥挤,但是bar呢?它也会被紧紧地打包在一起吗?__attribute__ ((__packed__))会使嵌套的struct也打包吗?
发布于 2013-06-04 03:43:45
不,bar不会被紧密打包。如果要打包,则必须将其显式标记为__attribute__ ((__packed__))。考虑以下示例:
#include <stdio.h>
struct
{
struct
{
char c;
int i;
} bar;
char c;
int i;
} foo1;
struct __attribute__ ((__packed__))
{
struct
{
char c;
int i;
} bar;
char c;
int i;
} foo2;
struct
{
struct __attribute__ ((__packed__))
{
char c;
int i;
} bar;
char c;
int i;
} foo3;
struct __attribute__ ((__packed__))
{
struct __attribute__ ((__packed__))
{
char c;
int i;
} bar;
char c;
int i;
} foo4;
int main()
{
printf("sizeof(foo1): %d\n", (int)sizeof(foo1));
printf("sizeof(foo2): %d\n", (int)sizeof(foo2));
printf("sizeof(foo3): %d\n", (int)sizeof(foo3));
printf("sizeof(foo4): %d\n", (int)sizeof(foo4));
return 0;
}本程序(使用64位的gcc 4.2和64位的clang 3.2编译)的输出为:
sizeof(foo1): 16
sizeof(foo2): 13
sizeof(foo3): 12
sizeof(foo4): 10如果要将struct及其嵌套的struct都紧密打包,则必须为每个struct显式声明__attribute__ ((__packed__))。如果您想要分离嵌套,以便在foo外部声明bar的类型,这是有意义的,如下所示:
// Note Bar is not packed
struct Bar
{
char c;
int i;
};
struct __attribute__ ((__packed__))
{
// Despite foo being packed, Bar is not, and thus bar will not be packed
struct Bar bar;
char c;
int i;
} foo;在上面的示例中,要打包bar,必须将Bar声明为__attribute__ ((__packed__))。如果您复制'n‘粘贴这些结构,以便像第一个代码示例中那样嵌套它们,您将看到打包行为是一致的。
对应的C++代码(使用g++ 4.2和clang++ 3.2编译,目标是64位,结果与上面完全相同):
#include <iostream>
struct Foo1
{
struct Bar1
{
char c;
int i;
} bar;
char c;
int i;
} foo1;
struct __attribute__ ((__packed__)) Foo2
{
struct Bar2
{
char c;
int i;
} bar;
char c;
int i;
} foo2;
struct Foo3
{
struct __attribute__ ((__packed__)) Bar3
{
char c;
int i;
} bar;
char c;
int i;
} foo3;
struct __attribute__ ((__packed__)) Foo4
{
struct __attribute__ ((__packed__)) Bar4
{
char c;
int i;
} bar;
char c;
int i;
} foo4;
int main()
{
std::cout << "sizeof(foo1): " << (int)sizeof(foo1) << std::endl;
std::cout << "sizeof(foo2): " << (int)sizeof(foo2) << std::endl;
std::cout << "sizeof(foo3): " << (int)sizeof(foo3) << std::endl;
std::cout << "sizeof(foo4): " << (int)sizeof(foo4) << std::endl;
}https://stackoverflow.com/questions/16904613
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