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

我只是想在这里添加我自己的解,因为我在任何地方都没有看到它。它是一个小而可移植的c++模板函数,并且只使用比特操作。

template<typename T> inline static T swapByteOrder(const T& val) {
    int totalBytes = sizeof(val);
    T swapped = (T) 0;
    for (int i = 0; i < totalBytes; ++i) {
        swapped |= (val >> (8*(totalBytes-i-1)) & 0xFF) << (8*i);
    }
    return swapped;
}

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。

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

如果您这样做是为了在不同平台之间传输数据,请查看ntoh和hton函数。

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);
}

这里有一个基本的函数来交换大小端序。它是基本的,但不需要补充库。

void endianness_swap(uint32_t& val) {
    uint8_t a, b, c;
    a = (val & 0xFF000000) >> 24;
    b = (val & 0x00FF0000) >> 16;
    c = (val & 0x0000FF00) >> 8;
    val=(val & 0x000000FF) << 24;
    val = val + (c << 16) + (b << 8) + (a);
}