双重奇怪的位腐败 [英] Strange bit corruption in a double
问题描述
我在双重中得到了一个非常奇怪的一点腐败。我在英特尔
Red Hat Linux上。 uname -a返回：
Linux foo.com 2.6.9-34.0.2.ELsmp＃1 SMP Fri Jun 30 10:33:58 EDT 2006
i686 i686 i386 GNU / Linux
我有一个双设置为0.00的变量。然后会出现一些数字
，然后，当我用printf（％f）打印这个变量
时，我得到0.00000。
然而，当我比较
如果（变量== 0.0），我得到假。
和if（变量0.0），我得到是的。
然后我运行了一个小函数来打印这个变量的位，并且
发现它的位模式很奇怪：
printf = 0.000000000000000
bits = 11001000 00010100 00010100 00001001 10001100 00000010 10111110
00000000
任何想法??????
FWIW，我知道打印双b / b
位模式的功能是正确的：
void print_binary_double（double value）
{
unsigned char * a;
a =（unsigned char *）& value ;
int bytes = sizeof（double）;
for（int i = 0; i< bytes; i ++）{
print_binary_uc（* a）;
printf（" ; ）;
a ++;
}
printf（" \ n"）;
}
void print_binary_uc（unsigned char value）
{
unsigned char value2;
int i;
int len = sizeof（unsigned char）* 8;
for（i = len-1; i> = 0; i--）
{
value2 =价值& （（unsigned char）1<< i）;
printf（"％d"，value2？1：0）;
}
}
I''m getting a very weird bit corruption in a double. I am on an Intel
Red Hat Linux box. uname -a returns:
Linux foo.com 2.6.9-34.0.2.ELsmp #1 SMP Fri Jun 30 10:33:58 EDT 2006
i686 i686 i386 GNU/Linux
I have a "double" variable that is set to 0.00. Some number
crunching then occurs, and later on, when I printf this variable
with printf("%f"), I am getting 0.00000.
However, when I compare
if (variable == 0.0), I get false.
and if (variable 0.0), I get true.
I then ran a small function to print the bits of this variable and
found that its bit pattern is quite odd:
printf = 0.000000000000000
bits = 11001000 00010100 00010100 00001001 10001100 00000010 10111110
00000000
Any ideas??????
FWIW, I know the function to print the bit pattern of the double
is correct:
void print_binary_double(double value)
{
unsigned char *a;
a = (unsigned char *)&value;
int bytes = sizeof(double);
for (int i = 0; i < bytes; i++) {
print_binary_uc(*a);
printf(" ");
a++;
}
printf("\n");
}
void print_binary_uc(unsigned char value)
{
unsigned char value2;
int i;
int len = sizeof(unsigned char) * 8;
for (i = len-1; i >= 0; i--)
{
value2 = value & ((unsigned char)1 << i);
printf("%d", value2 ? 1 : 0);
}
}
推荐答案
" Digital Puer" < di ********** @ hotmail.comwrote in message
news：11 ******************** **@v45g2000cwv.googlegr oups.com ...
"Digital Puer" <di**********@hotmail.comwrote in message
news:11**********************@v45g2000cwv.googlegr oups.com...
我在双重中遇到了非常奇怪的一点腐败。我在英特尔
Red Hat Linux上。 uname -a返回：
Linux foo.com 2.6.9-34.0.2.ELsmp＃1 SMP Fri Jun 30 10:33:58 EDT 2006
i686 i686 i386 GNU / Linux
我有一个双设置为0.00的变量。然后会出现一些数字
，然后，当我用printf（％f）打印这个变量
时，我得到0.00000。
然而，当我比较
如果（变量== 0.0），我得到假。
和if（变量0.0），我得到真正。
I''m getting a very weird bit corruption in a double. I am on an Intel
Red Hat Linux box. uname -a returns:
Linux foo.com 2.6.9-34.0.2.ELsmp #1 SMP Fri Jun 30 10:33:58 EDT 2006
i686 i686 i386 GNU/Linux
I have a "double" variable that is set to 0.00. Some number
crunching then occurs, and later on, when I printf this variable
with printf("%f"), I am getting 0.00000.
However, when I compare
if (variable == 0.0), I get false.
and if (variable 0.0), I get true.
我没有分析你提供的位模式，但你提供的信息是b
不一致腐败。
假设数字是正数，但非常小（让我们说10 ^（ - 30））。
然后没有版本的printf具有实际的小数位数将
显示除零之外的任何内容。另外，它会像你上面指出的那样测试为正面。
为了打印这个数字，你要使用％e ;而不是％f
格式说明符。
大多数二进制科学记数法（即浮点数，双精度）格式包含
二进制指数，我怀疑在这种情况下是一个非常负数。
数字中设置了很多1的事实并不矛盾
，正数非常小。
尝试％e，并发布结果......
I haven''t analyzed the bit pattern you provided, but the information you''ve
presented isn''t consistent with "corruption".
Assume that the number is positive, but very small (let''s say 10^(-30)).
Then no version of printf with a practical number of decimal places will
show anything but zero. Additionally, it would test as positive as you
indicated above.
In order to print this number, you''d to use the "%e" rather than the "%f"
format specifier.
Most binary scientific notation (i.e. float, double) formats contain a
binary exponent, which I suspect in this case is a very negative number.
The fact that there are a lot of "1"s set in the number are not inconsistent
with a very small positive number.
Try "%e", and post your results ...
" Digital Puer" < di ********** @ hotmail.comwrote in message
news：11 ******************** **@v45g2000cwv.googlegr oups.com ...
"Digital Puer" <di**********@hotmail.comwrote in message
news:11**********************@v45g2000cwv.googlegr oups.com...
我在双重中遇到了非常奇怪的一点腐败。我在英特尔
Red Hat Linux上。 uname -a返回：
Linux foo.com 2.6.9-34.0.2.ELsmp＃1 SMP Fri Jun 30 10:33:58 EDT 2006
i686 i686 i386 GNU / Linux
我有一个双设置为0.00的变量。然后会出现一些数字
，然后，当我用printf（％f）打印这个变量
时，我得到0.00000。
然而，当我比较
如果（变量== 0.0），我得到假。
和if（变量0.0），我得到是的。
然后我运行了一个小函数来打印这个变量的位，并且
发现它的位模式很奇怪：
printf = 0.000000000000000
bits = 11001000 00010100 00010100 00001001 10001100 00000010 10111110
00000000
任何想法??????
FWIW，我知道打印双b / b
位模式的功能是正确的：
void print_binary_double（double value）
{
unsigned char * a;
a =（unsigned char *）& value ;
int bytes = sizeof（double）;
for（int i = 0; i< bytes; i ++）{
print_binary_uc（* a）;
的printf（QUOT; ）;
a ++;
}
printf（" \ n"）;
}
void print_binary_uc（unsigned char value）
{
unsigned char value2;
int i;
int len = sizeof（unsigned char）* 8;
for（i = len-1; i> = 0; i--）
{
value2 =价值& （（unsigned char）1<< i）;
printf（"％d"，value2？1：0）;
}
}
I''m getting a very weird bit corruption in a double. I am on an Intel
Red Hat Linux box. uname -a returns:
Linux foo.com 2.6.9-34.0.2.ELsmp #1 SMP Fri Jun 30 10:33:58 EDT 2006
i686 i686 i386 GNU/Linux
I have a "double" variable that is set to 0.00. Some number
crunching then occurs, and later on, when I printf this variable
with printf("%f"), I am getting 0.00000.
However, when I compare
if (variable == 0.0), I get false.
and if (variable 0.0), I get true.
I then ran a small function to print the bits of this variable and
found that its bit pattern is quite odd:
printf = 0.000000000000000
bits = 11001000 00010100 00010100 00001001 10001100 00000010 10111110
00000000
Any ideas??????
FWIW, I know the function to print the bit pattern of the double
is correct:
void print_binary_double(double value)
{
unsigned char *a;
a = (unsigned char *)&value;
int bytes = sizeof(double);
for (int i = 0; i < bytes; i++) {
print_binary_uc(*a);
printf(" ");
a++;
}
printf("\n");
}
void print_binary_uc(unsigned char value)
{
unsigned char value2;
int i;
int len = sizeof(unsigned char) * 8;
for (i = len-1; i >= 0; i--)
{
value2 = value & ((unsigned char)1 << i);
printf("%d", value2 ? 1 : 0);
}
}
我敢打赌你的格式说明符需要tweeking。来源看起来很邋
。 LS
I''ll bet your format specifier needs tweeking. The source is sloppy-looking
too. LS
Lane Straatman说：
Lane Straatman said:
>
" Digital Puer" ; < di ********** @ hotmail.comwrote in message
news：11 ******************** **@v45g2000cwv.googlegr oups.com ...
>
"Digital Puer" <di**********@hotmail.comwrote in message
news:11**********************@v45g2000cwv.googlegr oups.com...
< snip>
<snip>
> void print_binary_double（double value）
{
unsigned char * a;
a =（unsigned char *）& value;
int bytes = sizeof（double）;
for（int i = 0; i< bytes; i ++）{
print_binary_uc（* a）;
printf（""）;
a ++;
}
printf（" \ n"）;
}
void print_binary_uc（unsigned char value）
{
unsigned char value2;
int i;
int len = sizeof（unsigned char）* 8;
for（i = len-1; i> = 0; i - ）
{
value2 = value& （（unsigned char）1<< i）;
printf（"％d"，value2？1：0）;
}
}
>void print_binary_double(double value)
{
unsigned char *a;
a = (unsigned char *)&value;
int bytes = sizeof(double);
for (int i = 0; i < bytes; i++) {
print_binary_uc(*a);
printf(" ");
a++;
}
printf("\n");
}
void print_binary_uc(unsigned char value)
{
unsigned char value2;
int i;
int len = sizeof(unsigned char) * 8;
for (i = len-1; i >= 0; i--)
{
value2 = value & ((unsigned char)1 << i);
printf("%d", value2 ? 1 : 0);
}
}
我敢打赌你的格式说明符需要tweeking。消息来源也很好看。 LS
I''ll bet your format specifier needs tweeking. The source is
sloppy-looking too. LS
您如何改善来源？
-
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在上述域名中， - www。
How would you improve the source?
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
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