reinterpret_cast - 将double解释为long [英] reinterpret_cast - to interpret double as long
问题描述
我想将double转换为二进制表示。我可以使用
"&"使用位掩码的位操作将*非*浮点类型转换为二进制
表示，但我不能使用&双打。
为了解决这个双倍的限制，我想保持
的两倍*相同*但改变它解释很久。我可以使用
"&"在多头。我试图使用reinterpret_cast来达到这个目的，但它每次都是
返回零。
double n1 = 32;
long n2 = *（reinterpret_cast< long *>（& n））; //返回零，对于所有n1
***如何将双倍的位保持为*相同*但将其
解释更改为long ？我做错了什么？
谢谢，
Suzanne
I''d like to convert a double to a binary representation. I can use the
"&" bit operation with a bit mask to convert *non* float types to binary
representations, but I can''t use "&" on doubles.
To get around this limitation on double, I''d like to keep the bits of
the double the *same* but change its interpretation to long. I can use
"&" on longs. I tried to use reinterpret_cast for this purpose, but it
returned zero every time.
double n1 = 32;
long n2 = *(reinterpret_cast<long*>(&n)); // returns zero, for all n1
*** How can I keep the bits of the double the *same* but change its
interpretation to long? What am I doing wrong?
Thanks,
Suzanne
推荐答案
你想要位模式还是值???
" Suzanne Vogel" <苏************* @ hotmail.com>在消息中写道
新闻：3f ********** @ news.unc.edu ...
Do you want the bit pattern or the value???
"Suzanne Vogel" <su*************@hotmail.com> wrote in message
news:3f**********@news.unc.edu...
我想转换一个双到二进制表示。我可以使用
"&"使用位掩码进行位操作以将*非*浮点类型转换为二进制
表示，但我不能使用&关于double的这个限制，我想保持
的两倍*相同*但是将它的解释改为long。我可以使用
"&"在多头。我试图为此目的使用reinterpret_cast，但它每次都返回零。
double n1 = 32;
long n2 = *（reinterpret_cast< long *>（& ; N））; //返回零，对于所有n1
***如何保持双倍的* *相同*但将其解释更改为long？我做错了什么？
谢谢，
Suzanne
I''d like to convert a double to a binary representation. I can use the
"&" bit operation with a bit mask to convert *non* float types to binary
representations, but I can''t use "&" on doubles.
To get around this limitation on double, I''d like to keep the bits of
the double the *same* but change its interpretation to long. I can use
"&" on longs. I tried to use reinterpret_cast for this purpose, but it
returned zero every time.
double n1 = 32;
long n2 = *(reinterpret_cast<long*>(&n)); // returns zero, for all n1
*** How can I keep the bits of the double the *same* but change its
interpretation to long? What am I doing wrong?
Thanks,
Suzanne
" Suzanne Vogel" <苏************* @ hotmail.com>写道...
"Suzanne Vogel" <su*************@hotmail.com> wrote...
我想将double转换为二进制表示。我可以使用
"&"使用位掩码进行位操作以将*非*浮点类型转换为二进制
表示，但我不能使用&关于double的这个限制，我想保持
的两倍*相同*但是将它的解释改为long。我可以使用
"&"在多头。我试图为此目的使用reinterpret_cast，但它每次都返回零。
double n1 = 32;
long n2 = *（reinterpret_cast< long *>（& ; N））; //返回零，对于所有n1
***如何保持双倍的* *相同*但将其解释更改为long？我做错了什么？
I''d like to convert a double to a binary representation. I can use the
"&" bit operation with a bit mask to convert *non* float types to binary
representations, but I can''t use "&" on doubles.
To get around this limitation on double, I''d like to keep the bits of
the double the *same* but change its interpretation to long. I can use
"&" on longs. I tried to use reinterpret_cast for this purpose, but it
returned zero every time.
double n1 = 32;
long n2 = *(reinterpret_cast<long*>(&n)); // returns zero, for all n1
*** How can I keep the bits of the double the *same* but change its
interpretation to long? What am I doing wrong?
在你的平台上''double''可能比''long''长。并且
您的硬件很可能是小端的。 ''32''作为一个双尾的
零尾数。这意味着只有UPPER两个字节有一些非零位。对于任何2的幂的双倍，低六个字节都是零
。当你要求双倍的
地址时，你会得到最低字节的地址。
将它重新解释为一个长的地址会给你四个较低的
字节。它们没有设置位。
你需要至少两个多头才能在平台上看到双b / b
的位模式。尝试这样做（非便携式！）：
double d1 = 32;
unsigned long * plong = reinterpret_cast< unsigned long *>（& d1）;
unsigned long n1 = plong [0]，n2 = plong [1];
并查看你在n1和n2中得到的结果。 （我使用unsigned long因为
它们对于位模式更好）
再一次：这段代码与标准C ++无关。它
产生_undefined_behaviour_。无论在
上做什么，你的平台都不是它应该在任何其他平台上做的事情
（甚至是同一编译器的不同版本）。
Victor
On your platform ''double'' is likely longer than ''long''. And
your hardware is likely little-endian. ''32'' as a double has
zero mantissa. That means that only the UPPER two bytes have
some non-zero bits in them. The lower six bytes are all zeroes
for any double that is a power of two. When you ask for the
address of the double, you get the address of the lowest byte.
Reinterpreting it as an address of a long gives you four lower
bytes. They have no bits set in them.
You need at least two longs to see the bit pattern in a double
on your platform. Try doing this (non-portable!):
double d1 = 32;
unsigned long *plong = reinterpret_cast<unsigned long*>(&d1);
unsigned long n1 = plong[0], n2 = plong[1];
and see what you get in n1 and n2. (I used unsigned long because
they are better for bit patterns)
Once again: this code has nothing to do with standard C++. It
produces _undefined_behaviour_. Whatever it happens to do on
your platform is not what it should do on any other platform
(or even with a different version of the same compiler).
Victor
" Suzanne Vogel" <苏************* @ hotmail.com>在消息中写道
新闻：3f ********** @ news.unc.edu ...
"Suzanne Vogel" <su*************@hotmail.com> wrote in message
news:3f**********@news.unc.edu...
我想转换一个双到二进制表示。我可以使用
"&"使用位掩码进行位操作以将*非*浮点类型转换为二进制
表示，但我不能使用&关于double的这个限制，我想保持
的两倍*相同*但是将它的解释改为long。我可以使用
"&"在多头。我试图为此目的使用reinterpret_cast，但它每次都返回零。
double n1 = 32;
long n2 = *（reinterpret_cast< long *>（& ; N））; //返回零，对于所有n1
***如何保持双倍的* *相同*但将其解释更改为long？我究竟做错了什么？
长度和double的大小差不多。我猜想
你只看到你的一半，而那一半都是零。
你可以试试这个
double n1 = 32;
long n2 = *（reinterpret_cast< long *>（& n））;
long n3 = *（reinterpret_cast< ; long *>（& n）+ 1）;
但更可靠的方法是使用工会
union
{
双d;
char c [sizeof（double）];
} conv;
conv.d = 32;
//以十六进制打印双倍
for（int i = 0; i< sizeof （double）; ++ i
cout<< hex<< setfill（''0''）<< setw（2）<< conv.c [i ];
谢谢，
Suzanne
I''d like to convert a double to a binary representation. I can use the
"&" bit operation with a bit mask to convert *non* float types to binary
representations, but I can''t use "&" on doubles.
To get around this limitation on double, I''d like to keep the bits of
the double the *same* but change its interpretation to long. I can use
"&" on longs. I tried to use reinterpret_cast for this purpose, but it
returned zero every time.
double n1 = 32;
long n2 = *(reinterpret_cast<long*>(&n)); // returns zero, for all n1
*** How can I keep the bits of the double the *same* but change its
interpretation to long? What am I doing wrong?
Its pretty unlikely that long is the same size as double. I would guess that
you are only seeing half your double, and that half is all zero.
You could try this
double n1 = 32;
long n2 = *(reinterpret_cast<long*>(&n));
long n3 = *(reinterpret_cast<long*>(&n) + 1);
But a more reliable method would be to use a union
union
{
double d;
char c[sizeof(double)];
} conv;
conv.d = 32;
// print out double in hex
for (int i =0; i < sizeof(double); ++i
cout << hex << setfill(''0'') << setw(2) << conv.c[i];
Thanks,
Suzanne
john
john
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