将float的位重新解释为unsigned long [英] Reinterpret the bits of a float as unsigned long
问题描述
对于哈希函数，我想将float
表达式的位重新解释为unsigned long。正常投射
（无符号长）float_expression
将浮点数截断为（无符号长整数）。
但这不是我想要的哈希函数的效果。
使用联合我的哈希函数可以实现：
union {
long unsigned hashvalue;
float floatvalue;
} value;
value.floatvalue = float_expression ;
现在
value.hashvalue
按预期包含重新解释的位。但我认为较短的
解决方案（没有赋值或memcpy变量）应该可以
。
当我演员首先到（void *）然后到（unsigned long）
gcc给我错误：
无法转换为指针类型
似乎gcc禁止向指针投射浮动。
有人有想法重新解释浮点的位置
unsigned long没有赋值或memcpy（某种类型
棘手的演员）。
问候Thomas Mertes
Seed7主页： http://seed7.sourceforge.net
维基百科： http://en.wikipedia.org/wiki/Seed7
项目页面： http://sourceforge.net/projects /秒eed7
< th *********** @ gmx.at>在消息中写道
新闻：11 ********************** @ g47g2000cwa.googlegr oups.com ...
对于哈希函数，我想将float
表达式的位重新解释为unsigned long。正常演员
（unsigned long）float_expression
将float缩减为（unsigned long）整数。
但这不是效果我想要一个哈希函数。
使用联合我的哈希函数可以实现：
union {
long unsigned hashvalue;
float floatvalue;
} value ;
value.floatvalue = float_expression;
现在
value.hashvalue
包含重新解释的位。但是我认为一个较短的解决方案（没有赋值或memcpy到变量）应该是可能的。
当我首先转换为（void *）并之后转换为（unsigned）很长）
gcc给我的错误：
无法转换为指针类型
gcc似乎禁止向指针投射浮动。
有没有人想要重新解释一个float的位，如果没有赋值或memcpy（某种类型的棘手演员），那就是无符号长。
如果float_expression是左值，你可以写：
*（unsigned long *）&（float_expression）
但这充满了危险。
PJ Plauger
Dinkumware，Ltd。
http://www.dinkumware.com
" PJ Plauger" <，P ... @ dinkumware.com>写道：
如果float_expression是一个左值，你可以写：
*（unsigned long *）&（float_expression）
>但这充满了危险。
感谢您的帮助。但是float_expression不是
被授予左值。你有一个rvalue解决方案吗？
问候Thomas Mertes
Seed7主页： http://seed7.sourceforge.net
维基百科： http://en.wikipedia.org/wiki/Seed7
项目页面： http://sourceforge.net/projects/seed7
th *********** @ gmx.at 写道：" PJ Plauger" <，P ... @ dinkumware.com>写道：
如果float_expression是一个左值，你可以写：
*（unsigned long *）&（float_expression）
>但这充满了危险。
感谢您的帮助。但是float_expression不被授予左值。你有一个rvalue解决方案吗？
你可以通过转换为
unsigned char *来获得基础表示，并通过它来完成。
unsigned char * raw =（unsigned char *）& floatval;
将raw [0]与raw [sizeof（float）-1]混合（例如构建一个
无符号长位......）
For a hash function I want to reinterpret the bits of a float
expression as unsigned long. The normal cast
(unsigned long) float_expression
truncates the float to an (unsigned long) integer.
But this is not the effect I want for a hash function.
With unions my hash function can be implemented:
union {
long unsigned hashvalue;
float floatvalue;
} value;
value.floatvalue = float_expression;
Now
value.hashvalue
contains the reinterpreted bits as intended. But I think a shorter
solution (without assignment or memcpy to a variable) should
be possible.
When I cast first to (void *) and after that to (unsigned long)
the gcc gives me the error:
cannot convert to a pointer type
It seems casting a float to a pointer is prohibited by gcc.
Has anybody an idea to reinterpret the bits of a float as
unsigned long without assignment or memcpy (some sort
of tricky cast).
Greetings Thomas Mertes
Seed7 Homepage: http://seed7.sourceforge.net
Wikipedia: http://en.wikipedia.org/wiki/Seed7
Project page: http://sourceforge.net/projects/seed7
<th***********@gmx.at> wrote in message
news:11**********************@g47g2000cwa.googlegr oups.com...
For a hash function I want to reinterpret the bits of a float
expression as unsigned long. The normal cast
(unsigned long) float_expression
truncates the float to an (unsigned long) integer.
But this is not the effect I want for a hash function.
With unions my hash function can be implemented:
union {
long unsigned hashvalue;
float floatvalue;
} value;
value.floatvalue = float_expression;
Now
value.hashvalue
contains the reinterpreted bits as intended. But I think a shorter
solution (without assignment or memcpy to a variable) should
be possible.
When I cast first to (void *) and after that to (unsigned long)
the gcc gives me the error:
cannot convert to a pointer type
It seems casting a float to a pointer is prohibited by gcc.
Has anybody an idea to reinterpret the bits of a float as
unsigned long without assignment or memcpy (some sort
of tricky cast).
If float_expression is a lvalue, you can write:
*(unsigned long *)&(float_expression)
but that is fraught with peril.
P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com
"P.J. Plauger" <p...@dinkumware.com> wrote:
If float_expression is a lvalue, you can write:
*(unsigned long *)&(float_expression)
but that is fraught with peril.
Thank you for your help. But the float_expression is not
granted to be an lvalue. Do you have an rvalue solution as well?
Greetings Thomas Mertes
Seed7 Homepage: http://seed7.sourceforge.net
Wikipedia: http://en.wikipedia.org/wiki/Seed7
Project page: http://sourceforge.net/projects/seed7
th***********@gmx.at wrote:"P.J. Plauger" <p...@dinkumware.com> wrote:If float_expression is a lvalue, you can write:
*(unsigned long *)&(float_expression)
but that is fraught with peril.
Thank you for your help. But the float_expression is not
granted to be an lvalue. Do you have an rvalue solution as well?
You can get at the underlying representation by casting to an
unsigned char* and work your way through that.
unsigned char *raw = (unsigned char *)&floatval;
Fiddle with raw[0] to raw[sizeof(float)-1] (e.g build an
unsigned long from the bits..)
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