malloc stumper [英] malloc stumper
问题描述
假设我有一些自定义类型foo，我知道定义为
a指向某些东西的指针，但我不知道是什么什么"是。
我想使用malloc为
foo类型的变量分配一些空间来指向。即我想使用malloc为
asome分配空间。并将malloc的返回值赋给一个变量
类型foo。但要做到这一点，malloc需要知道
某事的大小，而且我很难过。是否有任何方法可以使代码确定某事物的大小。来自foo？
谢谢，
jill
-
to s& e ^ n] d me m~a} i lr％e * m？o\v [有点来自我的| d）d：r {e：s] s。
Suppose that I have some custom type foo that I know is defined as
a "pointer to something", but I don''t know what "something" is.
I want to use malloc to allocate some space for a variable of type
foo to point to. I.e. I want to use malloc to allocate space for
a "something" and assign malloc''s return value to a variable of
type foo. But to do this, malloc needs to know the size of a
"something", and I''m stumped. Is there any way for the code to
determine the size of a "something" from foo?
Thanks,
jill
--
To s&e^n]d me m~a}i]l r%e*m?o\v[e bit from my a|d)d:r{e:s]s.
推荐答案
>假设我有一些自定义类型foo，我知道它被定义为
>Suppose that I have some custom type foo that I know is defined as
a指向什么，但我不知道什么是某事是。
如果你不知道什么是某事是，你怎么声明foo？
指向什么东西？ void *？
我想使用malloc为
foo类型的变量分配一些空间来指向。即我想使用malloc为
分配空间某事。并将malloc的返回值赋给
类型为foo的变量。但要做到这一点，malloc需要知道某事的大小，我很难过。有没有办法让代码确定某事物的大小？来自foo？
a "pointer to something", but I don''t know what "something" is.
If you don''t know what "something" is, how did you declare foo?
pointer to something? void *?
I want to use malloc to allocate some space for a variable of type
foo to point to. I.e. I want to use malloc to allocate space for
a "something" and assign malloc''s return value to a variable of
type foo. But to do this, malloc needs to know the size of a
"something", and I''m stumped. Is there any way for the code to
determine the size of a "something" from foo?
如果foo被声明为指向某事物的指针，你可以使用：
foo = malloc（sizeof （* foo））;
如果foo是void *或其他指针类型，这种方法将不起作用。
Gordon L. Burditt
If foo is declared as a pointer to something, you can use:
foo = malloc(sizeof(*foo));
If foo is a void * or other pointer type, this approach will not work.
Gordon L. Burditt
" J Krugman" < JK ********* @ yahbitoo.com>写了
"J Krugman" <jk*********@yahbitoo.com> wrote
假设我有一些自定义类型foo，我知道它被定义为指向某个东西的指针，但我不知道什么是某事。是。
我想使用malloc为
foo类型的变量分配一些空间来指向。
Suppose that I have some custom type foo that I know is defined as
a "pointer to something", but I don''t know what "something" is.
I want to use malloc to allocate some space for a variable of type
foo to point to.
typedef struct
{
int x;
} foo;
....
foo * fooptr = malloc（N * sizeof（foo））;
C没有的是类型变量，它包含一个在运行时确定的类型
。但是你通常可以使用
组合void * s和尺寸。假设我们想在
哈希表中存储一些foos。
typedef struct
{
size_t elsize;
void *条目;
} HASH;
HASH hashable（size_t element_size）
{
entries = malloc（element_size * TABLESIZE）;
elsize = element_size;
}
void store_entry（HASH * h，void * data，int（* hash）（void * data））
{
int slot = hash （数据）;
/ *我们需要一点复杂性以确保插槽是免费的，将滑过
这个* /
无符号char * ptr = h->条目;
memcpy（ptr + slot * h-> elsize，data，elsize）;
}
typedef struct
{
int x;
}foo;
....
foo *fooptr = malloc(N * sizeof(foo));
What C does not have is a "type" variable, that holds a type to be
determined at runtime. However you can usually get round this with a
combination of void *s and sizes. Let''s say we want to store some foos in a
hashtable.
typedef struct
{
size_t elsize;
void *entries;
} HASH;
HASH hashable(size_t element_size)
{
entries = malloc(element_size * TABLESIZE);
elsize = element_size;
}
void store_entry(HASH *h, void *data, int (*hash)(void *data))
{
int slot = hash(data);
/* we need a bit of complexity to make sure slot is free, will skate over
this */
unsigned char *ptr = h->entries;
memcpy(ptr + slot * h->elsize, data, elsize);
}
2004年12月4日星期六16:26:30 +0000（UTC）
J Krugman< jk ********* @ yahbitoo。 COM>写道：
On Sat, 4 Dec 2004 16:26:30 +0000 (UTC)
J Krugman <jk*********@yahbitoo.com> wrote:
假设我有一些自定义类型foo，我知道它被定义为
一个指向某个东西的指针，但我不知道是什么什么"是。
我想使用malloc为
foo类型的变量分配一些空间来指向。即我想使用malloc为
分配空间某事。并将malloc的返回值赋给
类型为foo的变量。但要做到这一点，malloc需要知道某事的大小，我很难过。有没有办法让代码确定某事物的大小？来自foo？
Suppose that I have some custom type foo that I know is defined as
a "pointer to something", but I don''t know what "something" is.
I want to use malloc to allocate some space for a variable of type
foo to point to. I.e. I want to use malloc to allocate space for
a "something" and assign malloc''s return value to a variable of
type foo. But to do this, malloc needs to know the size of a
"something", and I''m stumped. Is there any way for the code to
determine the size of a "something" from foo?
foo * bar = malloc（n * sizeof * bar）;
将为n分配空间bar类型的项目指向。这是
也很容易维护，因为你可以有
foo * bar;
/ *很多代码* /
bar = malloc（n * sizeof * bar）;
你不必找到定义栏的位置，看看malloc
是否足够分配空间。
-
Flash Gordon
生活在有趣的时代。
虽然我的电子邮件地址说垃圾邮件，它是真实的，我读了它。
foo *bar = malloc( n * sizeof *bar);
The will allocate space for n items of the type bar points to. It is
also easy to maintain since you could have
foo *bar;
/* lots of code */
bar = malloc( n * sizeof *bar);
and you don''t have to find where bar is defined to see if the malloc
allocates enough space.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.
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