有效的C ++？ [英] Valid C++?
问题描述
此代码是否在C ++下有效？我想使用这个结构，但我不是
确定它是否便携。
struct foo
{
char x [128];
};
struct bar
{
char sameSizeAsFooX [sizeof（（foo *）0） - > x];
};
谢谢，
Andrew
andrew queisser写道：这是代码低于有效的C ++？我想使用这个结构，但我不确定它是否可移动。
struct foo
{
char x [ 128];
};
结构栏
{char相同尺寸AsFooX [sizeof（（foo *）0） - > x];
};
是的，它是有效的。但是，我相信它也会导致未定义的
行为。
另一方面，它是一种常用的技术，尤其适用于C，用于
实例：
diff =（&（（foo *）0） - > x - （（foo *）0））
或类似的东西。
大多数编译器都会创建符合您期望的代码。由于你没有从x读取或写入它只是在
指针变量上进行算术运算。
andrew queisser写道：这段代码下面是否有效C ++？我想使用这个结构，但我不确定它是否可移动。
struct foo
{
char x [ 128];
};
结构栏
{char相同尺寸AsFooX [sizeof（（foo *）0） - > x];
};
虽然这似乎有用，但为了清楚起见，我会以不同的方式来做这件事。最糟糕的是，这样的事情：
char sameSizeAsFooX [sizeof foo（）。x];
更好，恕我直言，将是：
struct foo
{
enum {SIZE = 128};
char x [SIZE] ;
};
结构栏
{
char sameSizeAsFooX [foo :: SIZE] ;
};
干杯！ --M
andrew queisser写道：这段代码下面是否有效C ++？我想使用这个结构，但我不确定它是否可移动。
struct foo
{
char x [ 128];
};
结构栏
{char相同尺寸AsFooX [sizeof（（foo *）0） - > x];
我猜，没关系。看起来相当危险，比如解除引用
一个空指针。也许它不那么可怕了
char sameSizeAsFoox [sizeof foo（）。x];
（尽管确实需要' 'foo''是默认构造的，而
你的解决方案没有。）
};
V
-
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Is this code below valid C++? I''d like to use this construct but I''m not
sure if it''ll be portable.
struct foo
{
char x[128];
};
struct bar
{
char sameSizeAsFooX[ sizeof ((foo *)0)->x ];
};
Thanks,
Andrew
andrew queisser wrote:Is this code below valid C++? I''d like to use this construct but I''m not
sure if it''ll be portable.
struct foo
{
char x[128];
};
struct bar
{
char sameSizeAsFooX[ sizeof ((foo *)0)->x ];
};
Yes it is valid. However, I believe it also results in undefined
behavior.
On the other hand, it is a common technique to use especially in C, for
instance:
diff = (&((foo*)0)->x - ((foo*)0))
or something similar.
Most compilers create code that does what you expect here. Since you
are not reading from x or writing to it its just arith operations on a
pointer variable.
andrew queisser wrote:Is this code below valid C++? I''d like to use this construct but I''m not
sure if it''ll be portable.
struct foo
{
char x[128];
};
struct bar
{
char sameSizeAsFooX[ sizeof ((foo *)0)->x ];
};
Though that seems to work, for the sake of clarity, I would do it
differently. At worst, something like this:
char sameSizeAsFooX[ sizeof foo().x ];
Better, IMHO, would be:
struct foo
{
enum { SIZE = 128 };
char x[SIZE];
};
struct bar
{
char sameSizeAsFooX[ foo::SIZE ];
};
Cheers! --M
andrew queisser wrote:Is this code below valid C++? I''d like to use this construct but I''m
not sure if it''ll be portable.
struct foo
{
char x[128];
};
struct bar
{
char sameSizeAsFooX[ sizeof ((foo *)0)->x ];
It is OK, I guess. Seems rather dangerous though, like dereferencing
a null pointer. Perhaps it would be less scary to do
char sameSizeAsFoox[ sizeof foo().x ];
(although it does require for ''foo'' to be default-constructible while
your solution does not).
};
V
--
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I do not respond to top-posted replies, please don''t ask
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