填充和对齐 [英] Padding and alignment
问题描述
大家好，
我曾期望我的C ++编译器在类和结构中添加填充
将字段与其自然对齐边界。这似乎不是真的。
$ cat align.cxx
#include< cstdio>
struct foo
{
virtual void f（）const {}
long long x;
};
int main（）
{
foo bar;
printf（"％p％p \ n"，（void） *）& bar，（void *）& bar.x）;
返回0;
}
$ ./a.out
0xbfd9da00 0xbfd9da04
在我的平台上（x86 Linux g ++）long long是64位宽。
（我想前4个字节存储了v指针。）
为什么编译器没有添加填充以将x对齐到64位边界？
问候。
Hello everyone,
I had expected my C++ compiler to add padding within classes and structs
to align fields to their "natural" boundary. This seems not to be true.
$ cat align.cxx
#include <cstdio>
struct foo
{
virtual void f() const { }
long long x;
};
int main()
{
foo bar;
printf("%p %p\n", (void *)&bar, (void *)&bar.x);
return 0;
}
$ ./a.out
0xbfd9da00 0xbfd9da04
On my platform (x86 Linux g++) long long is 64-bits wide.
(I suppose the first 4 bytes store the v-pointer.)
Why didn''t the compiler add padding to align x to a 64-bit boundary?
Regards.
推荐答案
cat align.cxx
#include< cstdio>
struct foo
{
virtual void f（）const {}
long long x;
};
int main（）
{
foo bar;
printf（"％p％p \\\
&q uot;，（void *）& bar，（void *）& bar.x）;
返回0;
}
cat align.cxx
#include <cstdio>
struct foo
{
virtual void f() const { }
long long x;
};
int main()
{
foo bar;
printf("%p %p\n", (void *)&bar, (void *)&bar.x);
return 0;
}
./ a.out
0xbfd9da00 0xbfd9da04
在我的平台上（x86 Linux g ++）long long是64 - 比特宽。
（我想前4个字节存储了v指针。）
为什么编译器没有添加填充以将x对齐到64-位边界？
问候。
./a.out
0xbfd9da00 0xbfd9da04
On my platform (x86 Linux g++) long long is 64-bits wide.
(I suppose the first 4 bytes store the v-pointer.)
Why didn''t the compiler add padding to align x to a 64-bit boundary?
Regards.
Spoon写道：
Spoon wrote:
大家好，
我原本期望我的C ++编译器在类和结构中添加填充
将字段与其自然对齐边界。这似乎不是真的。
Hello everyone,
I had expected my C++ compiler to add padding within classes and structs
to align fields to their "natural" boundary. This seems not to be true.
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