指针在哪里指向？ [英] Where do pointers point to?
问题描述
是否有保证，指向C中任何对象的指针都指向该对象的最低寻址字节？
具体可能对于任何平台/ os /编译器组合：
（char *）v！=（void *）v
其中v是例如int变量。
如果是这样，任何现实生活中的例子？
" x-pander" <纳克*** @ pitek.eu.org>写道：
是否有保证，指向C中任何对象的指针恰好指向该对象的最低寻址字节？
是的，据我所知。 （我很想听到聪明的
对此有不同意见，因为我会从中学习。）
特定于任何平台/ os /编译器都可以组合：
（char *）v！=（void *）v
其中v是一个int变量，例如。
non sequitur ：你只是拖动整数到指针的转换，
这绝对是不便携的。
-
给我一个几年和一项大型研究补助金，
，我会给你一张收据。 - 理查德希思菲尔德
>具体来说，任何平台/ os /编译器组合都可以：（char *）v！=（void *）v
其中v是一个int变量，例如。
CORRECTION，当然（!!!）我的意思是：
（char *）& v！=（void *）& v
在文章< cs *********** @ mamut.aster.pl>，
" x-迎合" <纳克*** @ pitek.eu.org>写道：
是否有保证，指向C中任何对象的指针恰好位于该对象的最低寻址字节？
具体是否有可能为任何平台/ os /编译器组合：
（char *）v！=（void *）v
其中v是一个int变量，例如。
一个指向整个对象的指针。
如果你把一个指针转换成另一个类型，让我们说从double *到int *，
你得到一个指向一个对象的指针，该对象从与
第一个对象相同的位置开始。因此，如果你指向char *的指针，它指向对象的第一个
字节。
你无法比较char *和void *。在你的例子中，编译器将
自动将（void *）v转换为char *，所以你实际上是比较
$
（char *）v ==（char *）（void *）v
保证相等。
Is is guaranteed, that a pointer to any object in C points exactly at the
lowest addressed byte of this object?
Specifcally is it possible for any platform/os/compiler combination that:
(char *)v != (void *)v
where v is an int variable for example.
If so, any real-life examples?
"x-pander" <ng***@pitek.eu.org> writes:
Is is guaranteed, that a pointer to any object in C points exactly at the
lowest addressed byte of this object?
Yes, as far as I know. (I''d love to hear intelligent
disagreement on this, because I''d learn from it.)
Specifcally is it possible for any platform/os/compiler combination that:
(char *)v != (void *)v
where v is an int variable for example.
Non sequitur: you just dragged in integer-to-pointer conversion,
which is definitely non-portable.
--
"Give me a couple of years and a large research grant,
and I''ll give you a receipt." --Richard Heathfield
> Specifcally is it possible for any platform/os/compiler combination that:(char *)v != (void *)v
where v is an int variable for example.
CORRECTION, of course (!!!) i meant:
(char *)&v != (void *)&v
In article <cs***********@mamut.aster.pl>,
"x-pander" <ng***@pitek.eu.org> wrote:
Is is guaranteed, that a pointer to any object in C points exactly at the
lowest addressed byte of this object?
Specifcally is it possible for any platform/os/compiler combination that:
(char *)v != (void *)v
where v is an int variable for example.
A pointer points to the whole object.
If you cast a pointer to another type, lets say from double* to int*,
you get a pointer to an object that starts at the same place as the
first object. So if you cast a pointer to char*, it points to the first
byte of the object.
You cannot compare char* and void*. In your example, the compiler will
convert the (void *) v automatically to char*, so you are actually
comparing
(char *)v == (char *) (void *) v
which is guaranteed to be equal.
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