对象表示的位模式 [英] Bit-Pattern of Representation of Objects
问题描述
我正在努力想出一种方法来辨别C ++对象表示的实际
位模式（尤其是
小型内置类型的对象） ），我提出了
之后的混乱。但是我想知道，有没有更简单的方法
这样做？这看起来很笨拙。
#include< iostream>
#include< cmath>
template< typename T>
void Binary（T const& object）
{
size_t size = 8 * sizeof（object）; //以位为单位的对象大小。
unsigned long long int mask =
static_cast< unsigned long long int>
（pow（2.0， static_cast< double>（size - 1））+ 0.1）;
unsigned long long int pattern =
* reinterpret_cast< const unsigned long long int *>（& （）;（掩码0;掩码>> = 1）
$
else std :: cout<< 0;
}
std :: cout<< std :: endl;
返回;
}
-
干杯，
Robbie Hatley
美国加利福尼亚州东塔斯廷市
在pac bell dot net上孤独的狼内战
（put" [usenet]" ;受制于绕过垃圾邮件过滤器）
http://home.pacbell.net / earnur /
* Robbie Hatley：
我当时努力想出一种方法来辨别C ++对象表示的实际位模式（特别是
小内置类型的对象），我来了一塌糊涂之后的
。但是我想知道，有没有更简单的方法
这样做？这看起来很笨拙。
#include< iostream>
#include< cmath>
template< typename T>
void Binary（T const& object）
{
size_t size = 8 * sizeof（object）; //对象的大小（以位为单位）。
使用CHAR_BITS或其他任何名称：一个字节不一定是8位。
unsigned long long int mask =
C ++没有'long long'类型，但是。
static_cast< unsigned long long int>
（pow（2.0，static_cast< double>（size - 1））+ 0.1）;
使用左移运算符。
unsigned long long int pattern =
* reinterpret_cast< const unsigned long long int *>（& object）;
你不知道sizeopf（object）< = sizeof（long long）。
for（; mask 0; mask>> = 1）
{
if（pattern& mask）std :: cout< < " 1" ;;
else std :: cout<< " 0英寸;
std :: cout<< !!（模式和掩码）;
}
std :: cout<< std :: endl;
return;
不必要''返回''。
}
-
答：因为它弄乱了人们通常阅读文字的顺序。
问：为什么这么糟糕？
A：热门发布。
问：usenet和电子邮件中最烦人的事情是什么？
< blockquote> Robbie Hatley发布：
我正在努力想出一种方法来辨别C ++表示的实际
位模式对象
这里有一些我最近写的代码如果你感兴趣的话：
#include< cstddef> ;
#include< cassert>
#include< climits>
#include< ostream>
void PrintBits（void const * const mem，size_t amount_bytes，std :: ostream& os）
{
assert（mem）;
断言（ amount_bytes）;
char static str [CHAR_BIT + 1] = {};
unsigned char const * p = reinterpret_cast< unsigned char const *>（mem）;
do
{
unsigned const byte_val = * p ++;
char * pos = str;
unsigned to_and_with = 1U<< CHAR_BIT - 1;
do * pos ++ = byte_val& to_and_with？ ''1''：''0'';
while（to_and_with>> = 1）;
os<< str;
} while（--amount_bytes）;
}
模板< class T>
inline void PrintObjBits（T const& obj，std :: ostream& os）
{
PrintBits（& obj，sizeof obj，os）; < br $>
}
#include< iostream>
int main（）
{
long double array [4] = {241.126,632.225,2662.2523,23345.2352};
PrintObjBits（array，std :: cout）;
}
-
Frederick Gotham
Robbie Hatley写道：
我正在努力想出一种方法来辨别C ++对象表示的实际
位模式（特别是
小型内置类型的对象），然后我想出了
。但是我想知道，有没有更简单的方法
这样做？这看起来很笨拙。
将对象复制到一个大小合适的unsigned char数组中，然后转换为
十六进制的unsigned chars。如果你真的需要二进制而不是十六进制，
转换是一个相当简单的练习。
I was struggling to come up with a way to discern the actual
bit patterns of the representations of C++ objects (esp.
objects of small built-in types), and I came up with the
following mess. But I''m wondering, is there an easier way
to do this? This seems so klunky.
#include <iostream>
#include <cmath>
template<typename T>
void Binary(T const & object)
{
size_t size = 8 * sizeof(object); // Size of object in bits.
unsigned long long int mask =
static_cast<unsigned long long int>
(pow(2.0, static_cast<double>(size - 1)) + 0.1);
unsigned long long int pattern =
*reinterpret_cast<const unsigned long long int*>(&object);
for( ; mask 0 ; mask >>= 1)
{
if(pattern & mask) std::cout << "1";
else std::cout << "0";
}
std::cout << std::endl;
return;
}
--
Cheers,
Robbie Hatley
East Tustin, CA, USA
lone wolf intj at pac bell dot net
(put "[usenet]" in subject to bypass spam filter)
http://home.pacbell.net/earnur/
* Robbie Hatley:I was struggling to come up with a way to discern the actual
bit patterns of the representations of C++ objects (esp.
objects of small built-in types), and I came up with the
following mess. But I''m wondering, is there an easier way
to do this? This seems so klunky.
#include <iostream>
#include <cmath>
template<typename T>
void Binary(T const & object)
{
size_t size = 8 * sizeof(object); // Size of object in bits.Use CHAR_BITS or whatever it''s called: a byte isn''t necessarily 8 bits.
unsigned long long int mask =C++ does not have a ''long long'' type, yet.
static_cast<unsigned long long int>
(pow(2.0, static_cast<double>(size - 1)) + 0.1);Use left shift operator.
unsigned long long int pattern =
*reinterpret_cast<const unsigned long long int*>(&object);You don''t know that sizeopf(object) <= sizeof(long long).
for( ; mask 0 ; mask >>= 1)
{
if(pattern & mask) std::cout << "1";
else std::cout << "0";std::cout << !!(pattern & mask);
}
std::cout << std::endl;
return;Unnecessary ''return''.
}
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Robbie Hatley posted:
I was struggling to come up with a way to discern the actual
bit patterns of the representations of C++ objects
Here''s some code I wrote recently if you''re interested:
#include <cstddef>
#include <cassert>
#include <climits>
#include <ostream>
void PrintBits(void const * const mem,size_t amount_bytes,std::ostream &os)
{
assert( mem );
assert( amount_bytes );
char static str[CHAR_BIT + 1] = {};
unsigned char const *p = reinterpret_cast<unsigned char const*>(mem);
do
{
unsigned const byte_val = *p++;
char *pos = str;
unsigned to_and_with = 1U << CHAR_BIT - 1;
do *pos++ = byte_val & to_and_with ? ''1'' : ''0'';
while(to_and_with >>= 1);
os << str;
} while (--amount_bytes);
}
template<class T>
inline void PrintObjBits(T const &obj,std::ostream &os)
{
PrintBits(&obj,sizeof obj,os);
}
#include <iostream>
int main()
{
long double array[4] = { 241.126, 632.225, 2662.2523, 23345.2352 };
PrintObjBits(array,std::cout);
}
--
Frederick Gotham
Robbie Hatley wrote:I was struggling to come up with a way to discern the actual
bit patterns of the representations of C++ objects (esp.
objects of small built-in types), and I came up with the
following mess. But I''m wondering, is there an easier way
to do this? This seems so klunky.
Copy the object into a suitably sized array of unsigned char, then dump
the unsigned chars in hex. If you really need binary rather than hex,
the conversion is a fairly simple exercise.
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