Endian逆转？ [英] Endian Reversal?
问题描述
从BigEndian转换__int64类型的最快方法是什么？
Endian？
我写了这个函数：
---
uint64 EndianReverse（uint64 number）{
typedef unsigned __int8 uint8;
const int size = sizeof（uint64）;
uint8 polje [size] = {0};
uint8 * ptr = reinterpret_cast< uint8 *>（& number）;
int index = 0;
for（int i = size-1; i> = 0; i--）{
polje [index] = *（ptr + i）;
index ++;
}
uint64 * retPtr = reinterpret_cast< uint64 *>（polje）;
return * retPtr;
}
---
但是我知道有更快的逻辑和移位操作方式，所以如果有人
知道我会很感激。
What is the fastest way to convert __int64 type from BigEndian To Little
Endian?
I wrote this function:
---
uint64 EndianReverse(uint64 number) {
typedef unsigned __int8 uint8;
const int size = sizeof(uint64);
uint8 polje[size] = {0};
uint8 *ptr = reinterpret_cast<uint8*>(&number);
int index = 0;
for (int i=size-1 ; i>=0 ; i--) {
polje[index] = *(ptr+i);
index++;
}
uint64 *retPtr = reinterpret_cast<uint64*>(polje);
return *retPtr;
}
---
but i know there is faster way with logical and shift opertors, so if anyone
knows it i would be appreciated.
推荐答案
Tosha写道：
最快的方式是什么？将__int64类型从BigEndian转换为Little
Endian？
我写了这个有趣的ction：
---
uint64 EndianReverse（uint64 number）{
typedef unsigned __int8 uint8;
const int size = sizeof（uint64）;
uint8 polje [size ] = {0};
uint8 * ptr = reinterpret_cast< uint8 *>（& number）;
int index = 0;
for（int i = size-1; i> = 0; i--）{
polje [index] = *（ptr + i）;
index ++;
}
uint64 * retPtr = reinterpret_cast< uint64 *>（polje）;
返回* retPtr;
}
---
但我知道有更快的逻辑和移位操作方式，所以如果有人知道它我将不胜感激。
What is the fastest way to convert __int64 type from BigEndian To Little
Endian?
I wrote this function:
---
uint64 EndianReverse(uint64 number) {
typedef unsigned __int8 uint8;
const int size = sizeof(uint64);
uint8 polje[size] = {0};
uint8 *ptr = reinterpret_cast<uint8*>(&number);
int index = 0;
for (int i=size-1 ; i>=0 ; i--) {
polje[index] = *(ptr+i);
index++;
}
uint64 *retPtr = reinterpret_cast<uint64*>(polje);
return *retPtr;
}
---
but i know there is faster way with logical and shift opertors, so if anyone
knows it i would be appreciated.
牢度必须_measured_，而不是从使用的
操作的类型推断出来。也就是说，我希望这会更快一点：
模板< class T> T EndianReverse（T t）
{
unsigned char uc [sizeof t];
memcpy（uc，& t，sizeof t） ;
for（unsigned char * b = uc，* e = uc + sizeof（T） - 1; b< e; ++ b， - e）
std :: swap（* b，* e）;
memcpy（& t，uc，sizeof t）;
返回t;
}
....它在我的机器上（刚刚测量）大约20％。
V
-
请在邮寄回复时从我的地址删除资金
The "fastness" has to be _measured_, not deduced from the types of
operations used. That said, I''d expect this to be a bit faster:
template<class T> T EndianReverse(T t)
{
unsigned char uc[sizeof t];
memcpy(uc, &t, sizeof t);
for (unsigned char *b = uc, *e = uc + sizeof(T) - 1; b < e; ++b, --e)
std::swap(*b, *e);
memcpy(&t, uc, sizeof t);
return t;
}
.... and it is (just measured), by about 20%, on my machine.
V
--
Please remove capital As from my address when replying by mail
Tosha写道：
将__int64类型从BigEndian转换为Little的最快方法是什么？
What is the fastest way to convert __int64 type from BigEndian To Little
Endian?
< snip code>
既然你看起来像是在视觉工作室，那么尝试
_byteswap_uint64内在如何：
http://msdn2.microsoft.com/en-us/lib...77(VS。 80）的.aspx
<snip code>
Since it looks like you''re on visual studio, how about trying the
_byteswap_uint64 intrinsic:
http://msdn2.microsoft.com/en-us/lib...77(VS.80).aspx
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news:11**********************@p10g2000cwp.googlegr oups.com...
Tosha写道：
将__int64类型从BigEndian转换为Little的最快方法是什么？
What is the fastest way to convert __int64 type from BigEndian To Little
Endian?
< snip code>
因为看起来你在视觉工作室，尝试
_byteswap_uint64内在如何：
http://msdn2.microsoft.com/en-us/lib...77(VS.80 ）.aspx
非常感谢。
Thank you very much.
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