更改字节序的最快方法 [英] Quickest way to change endianness

问题描述

反转16位和32位整数的字节序的最快方法是什么。我通常会做类似的事情（此编码是在C ++中的Visual Studio中完成的）：

What is the quickest way to reverse the endianness of a 16 bit and 32 bit integer. I usually do something like (this coding was done in Visual Studio in C++):

union bytes4
{
    __int32 value;
    char ch[4];
};

union bytes2
{
    __int16 value;
    char ch[2];
};

__int16 changeEndianness16(__int16 val)
{
    bytes2 temp;
    temp.value=val;

    char x= temp.ch[0];
    temp.ch[0]=temp.ch[1];
    temp.ch[1]=x;
    return temp.value;
}

__int32 changeEndianness32(__int32 val)
{
    bytes4 temp;
    temp.value=val;
    char x;

    x= temp.ch[0];
    temp.ch[0]=temp.ch[1];
    temp.ch[1]=x;

    x= temp.ch[2];
    temp.ch[2]=temp.ch[3];
    temp.ch[3]=x;
    return temp.value;
}

有没有更快的方法，我不必进行很多计算？

Is there any faster way to do the same, in which I don't have to do so many calculations?

推荐答案

为什么不使用内置的 swab 函数，它可能比您的代码更好地优化了？

Why aren't you using the built-in swab function, which is likely optimized better than your code?

除此之外，通常的移位操作应该是快速开始，并且使用如此广泛，它们可能会被优化程序识别并被更好的代码取代。

Beyond that, the usual bit-shift operations should be fast to begin with, and are so widely used they may be recognized by the optimizer and replaced by even better code.

因为其他答案都有严重的错误，所以我将发布更好的实现：

Because other answers have serious bugs, I'll post a better implementation:

int16_t changeEndianness16(int16_t val)
{
    return (val << 8) |          // left-shift always fills with zeros
          ((val >> 8) & 0x00ff); // right-shift sign-extends, so force to zero
}

我测试过的编译器为此代码生成 rolw ，我认为稍长的序列（就指令数而言）实际上更快。基准会很有趣。

None of the compilers I tested generate rolw for this code, I think a slightly longer sequence (in terms of instruction count) is actually faster. Benchmarks would be interesting.

对于32位，有一些可能的操作命令：

For 32-bit, there are a few possible orders for the operations:

//version 1
int32_t changeEndianness32(int32_t val)
{
    return (val << 24) |
          ((val <<  8) & 0x00ff0000) |
          ((val >>  8) & 0x0000ff00) |
          ((val >> 24) & 0x000000ff);
}

//version 2, one less OR, but has data dependencies
int32_t changeEndianness32(int32_t val)
{
    int32_t tmp = (val << 16) |
                 ((val >> 16) & 0x00ffff);
    return ((tmp >> 8) & 0x00ff00ff) | ((tmp & 0x00ff00ff) << 8);
}

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