如何在c++中转换大端值和小端值?
为了清晰起见,我必须将二进制数据(双精度浮点值以及32位和64位整数)从一个CPU架构转换到另一个CPU架构。这并不涉及网络,因此ntoh()和类似的函数在这里不能工作。
注意:我接受的答案直接适用于我的目标编译器(这就是我选择它的原因)。然而,这里还有其他非常好的、更方便的答案。
如何在c++中转换大端值和小端值?
为了清晰起见,我必须将二进制数据(双精度浮点值以及32位和64位整数)从一个CPU架构转换到另一个CPU架构。这并不涉及网络,因此ntoh()和类似的函数在这里不能工作。
注意:我接受的答案直接适用于我的目标编译器(这就是我选择它的原因)。然而,这里还有其他非常好的、更方便的答案。
当前回答
下面介绍如何读取以IEEE 754 64位格式存储的double,即使您的主机使用不同的系统。
/*
* read a double from a stream in ieee754 format regardless of host
* encoding.
* fp - the stream
* bigendian - set to if big bytes first, clear for little bytes
* first
*
*/
double freadieee754(FILE *fp, int bigendian)
{
unsigned char buff[8];
int i;
double fnorm = 0.0;
unsigned char temp;
int sign;
int exponent;
double bitval;
int maski, mask;
int expbits = 11;
int significandbits = 52;
int shift;
double answer;
/* read the data */
for (i = 0; i < 8; i++)
buff[i] = fgetc(fp);
/* just reverse if not big-endian*/
if (!bigendian)
{
for (i = 0; i < 4; i++)
{
temp = buff[i];
buff[i] = buff[8 - i - 1];
buff[8 - i - 1] = temp;
}
}
sign = buff[0] & 0x80 ? -1 : 1;
/* exponet in raw format*/
exponent = ((buff[0] & 0x7F) << 4) | ((buff[1] & 0xF0) >> 4);
/* read inthe mantissa. Top bit is 0.5, the successive bits half*/
bitval = 0.5;
maski = 1;
mask = 0x08;
for (i = 0; i < significandbits; i++)
{
if (buff[maski] & mask)
fnorm += bitval;
bitval /= 2.0;
mask >>= 1;
if (mask == 0)
{
mask = 0x80;
maski++;
}
}
/* handle zero specially */
if (exponent == 0 && fnorm == 0)
return 0.0;
shift = exponent - ((1 << (expbits - 1)) - 1); /* exponent = shift + bias */
/* nans have exp 1024 and non-zero mantissa */
if (shift == 1024 && fnorm != 0)
return sqrt(-1.0);
/*infinity*/
if (shift == 1024 && fnorm == 0)
{
#ifdef INFINITY
return sign == 1 ? INFINITY : -INFINITY;
#endif
return (sign * 1.0) / 0.0;
}
if (shift > -1023)
{
answer = ldexp(fnorm + 1.0, shift);
return answer * sign;
}
else
{
/* denormalised numbers */
if (fnorm == 0.0)
return 0.0;
shift = -1022;
while (fnorm < 1.0)
{
fnorm *= 2;
shift--;
}
answer = ldexp(fnorm, shift);
return answer * sign;
}
}
对于这套函数的其余部分,包括写和整数例程,请参阅我的github项目
https://github.com/MalcolmMcLean/ieee754
其他回答
在大多数POSIX系统中(虽然不是在POSIX标准中)有end .h,它可以用来确定系统使用的编码。然后是这样的:
unsigned int change_endian(unsigned int x)
{
unsigned char *ptr = (unsigned char *)&x;
return (ptr[0] << 24) | (ptr[1] << 16) | (ptr[2] << 8) | ptr[3];
}
这将交换顺序(从大端序到小端序):
如果你有数字0xDEADBEEF(在一个小端序系统中存储为0xEFBEADDE), ptr[0]将是0xEF, ptr[1]是0xBE,等等。
但是如果你想将它用于网络,那么htons, htonl和htonll(以及它们的逆ntohs, ntohl和ntohll)将有助于从主机顺序转换到网络顺序。
c++20无分支版本,现在std::endian已经存在,但在c++23之前增加了std::byteswap
#include <bit>
#include <type_traits>
#include <concepts>
#include <array>
#include <cstring>
#include <iostream>
#include <bitset>
template <int LEN, int OFF=LEN/2>
class do_swap
{
// FOR 8 bytes:
// LEN=8 (LEN/2==4) <H><G><F><E><D><C><B><A>
// OFF=4: FROM=0, TO=7 => [A]<G><F><E><D><C><B>[H]
// OFF=3: FROM=1, TO=6 => [A][B]<F><E><D><C>[G][H]
// OFF=2: FROM=2, TO=5 => [A][B][C]<E><D>[F][G][H]
// OFF=1: FROM=3, TO=4 => [A][B][C][D][E][F][G][H]
// OFF=0: FROM=4, TO=3 => DONE
public:
enum consts {FROM=LEN/2-OFF, TO=(LEN-1)-FROM};
using NXT=do_swap<LEN, OFF-1>;
// flip the first and last for the current iteration's range
static void flip(std::array<std::byte, LEN>& b)
{
std::byte tmp=b[FROM];
b[FROM]=b[TO];
b[TO]=tmp;
NXT::flip(b);
}
};
template <int LEN>
class do_swap<LEN, 0> // STOP the template recursion
{
public:
static void flip(std::array<std::byte, LEN>&)
{
}
};
template<std::integral T, std::endian TO, std::endian FROM=std::endian::native>
requires ((TO==std::endian::big) || (TO==std::endian::little))
&& ((FROM==std::endian::big) || (FROM==std::endian::little))
class endian_swap
{
public:
enum consts {BYTE_COUNT=sizeof(T)};
static T cvt(const T integral)
{
// if FROM and TO are the same -- nothing to do
if (TO==FROM)
{
return integral;
}
// endian::big --> endian::little is the same as endian::little --> endian::big
// the bytes have to be reversed
// memcpy seems to be the most supported way to do byte swaps in a defined way
std::array<std::byte, BYTE_COUNT> bytes;
std::memcpy(&bytes, &integral, BYTE_COUNT);
do_swap<BYTE_COUNT>::flip(bytes);
T ret;
std::memcpy(&ret, &bytes, BYTE_COUNT);
return ret;
}
};
std::endian big()
{
return std::endian::big;
}
std::endian little()
{
return std::endian::little;
}
std::endian native()
{
return std::endian::native;
}
long long swap_to_big(long long x)
{
return endian_swap<long long, std::endian::big>::cvt(x);
}
long long swap_to_little(long long x)
{
return endian_swap<long long, std::endian::little>::cvt(x);
}
void show(std::string label, long long x)
{
std::cout << label << "\t: " << std::bitset<64>(x) << " (" << x << ")" << std::endl;
}
int main(int argv, char ** argc)
{
long long init=0xF8FCFEFF7F3F1F0;
long long to_big=swap_to_big(init);
long long to_little=swap_to_little(init);
show("Init", init);
show(">big", to_big);
show(">little", to_little);
}
使用下面的代码,您可以轻松地在BigEndian和LittleEndian之间进行切换
#define uint32_t unsigned
#define uint16_t unsigned short
#define swap16(x) ((((uint16_t)(x) & 0x00ff)<<8)| \
(((uint16_t)(x) & 0xff00)>>8))
#define swap32(x) ((((uint32_t)(x) & 0x000000ff)<<24)| \
(((uint32_t)(x) & 0x0000ff00)<<8)| \
(((uint32_t)(x) & 0x00ff0000)>>8)| \
(((uint32_t)(x) & 0xff000000)>>24))
虽然没有使用固有函数有效,但肯定是可移植的。我的回答:
#include <cstdint>
#include <type_traits>
/**
* Perform an endian swap of bytes against a templatized unsigned word.
*
* @tparam value_type The data type to perform the endian swap against.
* @param value The data value to swap.
*
* @return value_type The resulting swapped word.
*/
template <typename value_type>
constexpr inline auto endian_swap(value_type value) -> value_type
{
using half_type = typename std::conditional<
sizeof(value_type) == 8u,
uint32_t,
typename std::conditional<sizeof(value_type) == 4u, uint16_t, uint8_t>::
type>::type;
size_t const half_bits = sizeof(value_type) * 8u / 2u;
half_type const upper_half = static_cast<half_type>(value >> half_bits);
half_type const lower_half = static_cast<half_type>(value);
if (sizeof(value_type) == 2u)
{
return (static_cast<value_type>(lower_half) << half_bits) | upper_half;
}
return ((static_cast<value_type>(endian_swap(lower_half)) << half_bits) |
endian_swap(upper_half));
}
我们已经用模板做到了这一点。你可以这样做:
// Specialization for 2-byte types.
template<>
inline void endian_byte_swapper< 2 >(char* dest, char const* src)
{
// Use bit manipulations instead of accessing individual bytes from memory, much faster.
ushort* p_dest = reinterpret_cast< ushort* >(dest);
ushort const* const p_src = reinterpret_cast< ushort const* >(src);
*p_dest = (*p_src >> 8) | (*p_src << 8);
}
// Specialization for 4-byte types.
template<>
inline void endian_byte_swapper< 4 >(char* dest, char const* src)
{
// Use bit manipulations instead of accessing individual bytes from memory, much faster.
uint* p_dest = reinterpret_cast< uint* >(dest);
uint const* const p_src = reinterpret_cast< uint const* >(src);
*p_dest = (*p_src >> 24) | ((*p_src & 0x00ff0000) >> 8) | ((*p_src & 0x0000ff00) << 8) | (*p_src << 24);
}