void vs void *（哲学问题） [英] void vs void* (philosophical question)
问题描述
根据标准（ISO C99草案WG14 / N1124），void是无法完成的
不完整类型，包含空的
值。既然它声明了一个值的缺席可以被认为是一个
数据类型？
关于void *，它只是简单地重用相同的关键字（无效）或
他们有更密切的关系？我只想到一个 - 一个指针
到一个不完整的类型。但是，如果有人可以说虚空是一个值的价值，我们怎么能指向一个不存在
的东西呢？
-
一个人的自由停止在别人'开始的地方
Giannis Papadopoulos
计算机与通信工程部。 （CCED）
塞萨利大学
http：// dop .freegr.net /
According to the standard (ISO C99 draft WG14/N1124), void is the
incomplete type that cannot be completed and comprises the empty set of
values. Since it declares the absense of a value can it be considered a
data type?
Regarding void*, is it just a simple reuse of the same keyword (void) or
they have a closer relationship? I could think of only one - a pointer
to an incomplete type. However, if one could say that void is the
absense of a value, how can we have a pointer to something that does not
exist?
--
one''s freedom stops where others'' begin
Giannis Papadopoulos
Computer and Communications Engineering dept. (CCED)
University of Thessaly
http://dop.freegr.net/
推荐答案
Giannis Papadopoulos写道：
Giannis Papadopoulos wrote:
根据标准（ISO C99草案） WG14 / N1124），void是无法完成的
不完整类型，包含空的
值。既然它声明了某个值的缺失可以被认为是一种数据类型吗？
关于void *，它只是简单地重用相同的关键字（void）或
他们有更密切的关系？我只想到一个 - 指向不完整类型的指针。但是，如果有人可以说虚空是一个没有价值的东西，我们怎么能指向一个不存在的东西呢？
According to the standard (ISO C99 draft WG14/N1124), void is the
incomplete type that cannot be completed and comprises the empty set of
values. Since it declares the absense of a value can it be considered a
data type?
Regarding void*, is it just a simple reuse of the same keyword (void) or
they have a closer relationship? I could think of only one - a pointer
to an incomplete type. However, if one could say that void is the
absense of a value, how can we have a pointer to something that does not
exist?
这是'我最好的解释：
void *是指向（几乎）任何东西的指针。如果你取消引用
void *，你可能指的是32位整数或128位
浮点值。你可能指的是一个结构或一个
联盟。你可能指的是该死的__anything__。在C中，你做了簿记，而不是编译器（通常是）。
因为无效*可能指向任何东西，它是'最安全的
编译器将其视为特别指向__nothing的情况。你需要在使用之前将$ *
转换为另一种指针类型：
#include< stdio.h>
int main（无效）
{
int a = 42;
void * p =& a;
printf（"％d \ n"，*（int *）p）;
返回0;
}
[mark @ icepick]
Here''s my best explanation:
A void * is a pointer to (just about) anything. If you dereference a
void *, you might be referring to a 32 bit integer or a 128 bit
floating point value. You might be referring to a structure or a
union. You might be referring to damn near __anything__. In C, you do
the bookkeeping, not the compiler (usually).
Since a void * may point to just about anything, it''s safest for the
compiler to treat it as if it points to __nothing in particular__. You
must convert a void * to another pointer type before using it:
#include <stdio.h>
int main(void)
{
int a = 42;
void *p = &a;
printf("%d\n", *(int *) p);
return 0;
}
[mark@icepick]
gcc -ansi -pedantic -Wall -O2 -o foo foo.c
[mark @ icepick]
gcc -ansi -pedantic -Wall -O2 -o foo foo.c
[mark@icepick]
./ foo
42
Mark F. Haigh
mf ***** @ sbcglobal.net
./foo
42
Mark F. Haigh
mf*****@sbcglobal.net
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