类型转换void指针？ [英] Type-casting void pointers?
问题描述
Obrhy / hugo 2004年7月12日
是否有一个很好的理由无效指针
args被转换为byte只是为了将它们的
地址分配给字节指针？
/ *来自Steve Maguire的''编写代码''* /
void * memcpy（void * pvTo，void * pvFrom，size_t size）
{
byte * pbTo =（byte *）pvTo;
byte * pbFrom =（byte *）pvFrom;
while（size--> 0）
* pbTo ++ = * pbFrom ++;
返回（pvTo）;
}
地址都是unsigned int。为什么不
只需字节* pbTo = pvTo;初始化？
hugo ---------
Obrhy/hugo July 12, 2004
Take a look at this memcpy() definition.
Is there a good reason the void pointer
args are cast to byte just to assign their
addresses to byte pointers?
/*from Steve Maguire''s ''Writing Soild Code''*/
void *memcpy(void *pvTo,void *pvFrom,size_t size)
{
byte *pbTo = (byte *)pvTo;
byte *pbFrom = (byte *)pvFrom;
while(size-- >0)
*pbTo++ = *pbFrom++;
return (pvTo);
}
The addresses are all unsigned int. Why not
simply byte *pbTo = pvTo; to initialize?
hugo ---------
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Obrhy / hugo 2004年7月12日
看看这个memcpy（）定义。
是否有一个很好的理由将void指针
args强制转换为字节将他们的
地址分配给字节指针？
/ *来自Steve Maguire'''Writing Soild Code''* /
void * memcpy（void * pvTo，void * pvFrom，size_t size）
{
byte * pbTo =（byte *）pvTo;
byte * pbFrom =（byte *）pvFrom;
while（size - > 0）
* pbTo ++ = * pbFrom ++;
返回（pvTo）;
}
地址都是unsigned int。为什么不
简单的字节* pbTo = pvTo;初始化？
Obrhy/hugo July 12, 2004
Take a look at this memcpy() definition.
Is there a good reason the void pointer
args are cast to byte just to assign their
addresses to byte pointers?
/*from Steve Maguire''s ''Writing Soild Code''*/
void *memcpy(void *pvTo,void *pvFrom,size_t size)
{
byte *pbTo = (byte *)pvTo;
byte *pbFrom = (byte *)pvFrom;
while(size-- >0)
*pbTo++ = *pbFrom++;
return (pvTo);
}
The addresses are all unsigned int. Why not
simply byte *pbTo = pvTo; to initialize?
IMO没有必要，如果这些无效*。
除非有人的Clib实现打破了，size_t是一个非负的
类型，因此不需要放大小 - > 0，只需要大小就可以了。
我在某处看过ssize_t linux最近，甚至在单一的unix规范中。
愚蠢的东西IMO。它限制了有效范围，允许签名值询问
的问题...
Alex
IMO there''s no need in this if those are void*.
And unless someone''s Clib implementation is broken, size_t is a nonnegative
type, hence no need to put size-->0, simply size would do.
I''ve seen ssize_t somewhere in linux recently, even in the single unix spec.
Stupid thing IMO. It limits the valid range by allowing signed values asking
for problems...
Alex
hugo2写道：
Obrhy / hugo 2004年7月12日
看看这个memcpy（）定义。
有没有一个很好的理由，无效指针
args被转换为byte只是为了将它们的
地址分配给字节指针？
/ *来自Steve Maguire'''Writing Soild Code''* /
您是否忘记了''e'或错误的''我'
和'l''的顺序？ SCNR
void * memcpy（void * pvTo，void * pvFrom，size_t size）
{
byte * pbTo =（byte *）pvTo;
byte * pbFrom =（byte *）pvFrom;
while（size--> 0）
* pbTo ++ = * pbFrom ++;
返回（pvTo）;
}
地址都是unsigned int。为什么不
简单的字节* pbTo = pvTo;初始化？
Obrhy/hugo July 12, 2004
Take a look at this memcpy() definition.
Is there a good reason the void pointer
args are cast to byte just to assign their
addresses to byte pointers?
/*from Steve Maguire''s ''Writing Soild Code''*/
Did you forget an ''e'' or get wrong the order of ''i''
and ''l''? SCNR
void *memcpy(void *pvTo,void *pvFrom,size_t size)
{
byte *pbTo = (byte *)pvTo;
byte *pbFrom = (byte *)pvFrom;
while(size-- >0)
*pbTo++ = *pbFrom++;
return (pvTo);
}
The addresses are all unsigned int. Why not
simply byte *pbTo = pvTo; to initialize?
在void *的情况下，强制转换是完全没必要的。
我不知道类型的定义;字节"但是如果
它不是unsigned char的typedef，我会
建议您看看
clc而不是相信这本书：不必要的演员
是一件坏事。
干杯
迈克尔
-
电子邮件：我的是/ at / gmx / dot / de地址。
In the case of void*, the cast is completely unnecessary.
I do not know the definition of the type "byte" but if
it is anything other than a typedef for "unsigned char", I would
suggest that you have a look at the wisdom to be found in
c.l.c rather than believing this book: Unnecessary casts
are a Bad Thing.
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.
hugo2写道：
hugo2 wrote:
Obrhy / hugo 2004年7月12日
看一下这个memcpy（）的定义。
这是否有充分理由指针
args被转换为字节只是为了将它们的地址分配给字节指针？
/ *来自Steve Maguire'''Writing Soild Code''* /
void * memcpy（void * pvTo，void * pvFrom，size_t size）
{
byte * pbTo =（byte *）pvTo;
byte * pbFrom =（byte *）pvFrom;
while（size--> 0）
* pbTo ++ = * pbFrom ++;
return（pvTo）;
}
< br / >地址都是unsigned int。为什么不
简单的字节* pbTo = pvTo;初始化？
Obrhy/hugo July 12, 2004
Take a look at this memcpy() definition.
Is there a good reason the void pointer
args are cast to byte just to assign their
addresses to byte pointers?
/*from Steve Maguire''s ''Writing Soild Code''*/
void *memcpy(void *pvTo,void *pvFrom,size_t size)
{
byte *pbTo = (byte *)pvTo;
byte *pbFrom = (byte *)pvFrom;
while(size-- >0)
*pbTo++ = *pbFrom++;
return (pvTo);
}
The addresses are all unsigned int. Why not
simply byte *pbTo = pvTo; to initialize?
你是对的，C中不需要强制转换。这将是C ++，所以
可能是这种用法的原因。它类似于来自* alloc（）系列的
返回的强制类型转换，不是必需的，但是很多代码
无论如何都要执行它。我对这本书一无所知
你提到的。
Brian
You are correct, the cast is not required in C. It would be in C++, so
that could be reason for that usage. It''s similar to the casts for the
returns from the *alloc() family, not necessary but a LOT of code
examples out there do it anyway. I don''t know anything about that book
you mention.
Brian
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