将void *转换为void **？ [英] Casting void * to void ** ?
问题描述
大家好，
我有一个问题听起来很基本。
我的结构很简单：
struct simple {
void * buffer;
};
typedef struct simple简单;
在我的函数中我这样做：
void do_Something（）{
Simple * simp_struct;
simp_struct-> buffer = malloc（10 * sizeof（int *））;
call_func（（void **）（（int **）（simp_struct - >缓冲区）））;
....
}
函数call_func有这个原型：
call_func（void ** buf）;
我对这段代码感到困惑：
call_func（（void **） （（int **）（simp_struct-> buffer）））;
这个结构是什么意思？如何将simp_struct->缓冲区
（这是一个void *）强制转换为int **，然后将
强制转换为void **并传递给call_func？
Rgds。
Mirage
Twister写道：< blockquote class =post_quotes>大家好，
我有一个听起来很基本的问题。
我的结构很简单：
struct simple {
void * buffer;
};
typedef struct simple简单;
在我的函数中我这样做：
void do_Something（）{
简单* simp_struct;
simp_struct-> buffer = malloc（10 * sizeof（int *））;
call_func（（void * *）（（int **）（simp_struct-> buffer）））;
....
函数call_func有这个原型：
call_func（void ** buf）;
我对这段代码很困惑：
call_func（（void **）（（int **）（simp_struct-> buffer）） ）;
这是什么缺点结合意味着什么？如何将simp_struct->缓冲区
（这是一个void *）强制转换为int **，后跟
强制转换为void **并传递给call_func？
Rgds。
幻影
我错误输入了以前邮件的一部分：
这段代码：我很困惑这段代码：
call_func（（void **）（（int **）（simp_struct-> buffer）））;
应该是这个：
for（i = 0; i< 10; i ++）
call_func（（void **）（（int **）simp_struct-> ;缓冲区+ i））;
我的问题依然如故。以上
构造的含义是什么意思？
Rgds。
Mirage
Twister说：
< snip>
simp_struct-> buffer = malloc（10 * sizeof（int *）） ; call_func（（void **）（（int **）（simp_struct-> buffer）））;
为什么不这样做：
call_func（& simp_struct-> buffer）;
强制转换几乎总是错误的。
函数call_func有这个原型：
call_func（void ** buf）;
我是与这段代码混淆：
call_func（（void **）（（int **）（simp_struct-> buffer）））;
这个构造意味着什么？
这意味着你不会（或者写不出来的人）了解什么铸件
for。
< br $>
-
Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上面的域名（但显然放弃了www）
在文章< k8 **** *********@news.oracle.com>，
Twister< tw ***** @ nospam.com>写道：
struct simple {
void * buffer;
};
typedef struct simple Simple;
void do_Something（）{
Simple * simp_struct;
simp_struct是该语句后的未初始化指针。
simp_struct-> buffer = malloc（10 * sizeof（int *））;
但是你正在使用它，就像它被初始化一样。
simp_struct->缓冲区首先解除引用simp_struct和
然后在那里访问名为buffer的结构组件，所以
首先需要给simp_struct一个值。
call_func（（void **）（（int **）（simp_struct） - >缓冲区）））;
....
}
-
原型是其克隆的超类型。 - maplesoft
Hi All,
I have a question which might sound very basic.
I have a simple structure:
struct simple{
void *buffer;
};
typedef struct simple Simple;
In my function I do this:
void do_Something(){
Simple *simp_struct;
simp_struct->buffer = malloc(10 * sizeof(int *));
call_func((void **)((int **)(simp_struct->buffer)));
....
}
The function call_func has this prototype:
call_func(void **buf);
I am confused with this piece of code:
call_func((void **)((int **)(simp_struct->buffer)));
What does this construct mean? How is that simp_struct->buffer
(which is a void *) is being cast to a int** followed by a
cast to void ** and passed to call_func ?
Rgds.
Mirage
Twister wrote:Hi All,
I have a question which might sound very basic.
I have a simple structure:
struct simple{
void *buffer;
};
typedef struct simple Simple;
In my function I do this:
void do_Something(){
Simple *simp_struct;
simp_struct->buffer = malloc(10 * sizeof(int *));
call_func((void **)((int **)(simp_struct->buffer)));
....
}
The function call_func has this prototype:
call_func(void **buf);
I am confused with this piece of code:
call_func((void **)((int **)(simp_struct->buffer)));
What does this construct mean? How is that simp_struct->buffer
(which is a void *) is being cast to a int** followed by a
cast to void ** and passed to call_func ?
Rgds.
Mirage
I mistyped part of my previous mail:
This piece of code: I am confused with this piece of code:
call_func((void **)((int **)(simp_struct->buffer)));
should be this:
for(i=0; i<10 ;i++)
call_func((void **)((int **)simp_struct->buffer + i));
My question remains the same. What does the above
construct mean?
Rgds.
Mirage
Twister said:
<snip>
simp_struct->buffer = malloc(10 * sizeof(int *));
call_func((void **)((int **)(simp_struct->buffer)));
Why not just do this:
call_func(&simp_struct->buffer);
Casts are almost always wrong.
The function call_func has this prototype:
call_func(void **buf);
I am confused with this piece of code:
call_func((void **)((int **)(simp_struct->buffer)));
What does this construct mean?
It means you don''t (or whoever wrote it doesn''t) understand what casting is
for.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
In article <k8*************@news.oracle.com>,
Twister <tw*****@nospam.com> wrote:
struct simple{
void *buffer;
};
typedef struct simple Simple; void do_Something(){
Simple *simp_struct;
simp_struct is an uninitialized pointer after that statement.
simp_struct->buffer = malloc(10 * sizeof(int *));
But there you are using it as if it was initialized.
simp_struct->buffer involves dereferencing simp_struct first and
then accessing the structure component named buffer there, so
simp_struct needs to be given a value first.
call_func((void **)((int **)(simp_struct->buffer)));
....
}
--
Prototypes are supertypes of their clones. -- maplesoft
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