快速 log2(float x) 实现 C++ [英] Fast log2(float x) implementation C++

问题描述

我需要在 C++ 中非常快速地实现 log2(float x) 函数.

I need a Very-Fast implementation of log2(float x) function in C++.

我发现了一个非常有趣的实现(而且速度非常快！)

I found a very interesting implementation (and extremely fast!)

#include <intrin.h>

inline unsigned long log2(int x)
{
    unsigned long y;
    _BitScanReverse(&y, x);
    return y;
}

但是这个函数只适用于输入中的整数值.

But this function is good only for integer values in input.

问题:有没有办法把这个函数转换成double类型的输入变量?

Question: Is there any way to convert this function to double type input variable?

UPD:

我找到了这个实现:

typedef unsigned long uint32;
typedef long int32;   
static inline int32 ilog2(float x)
{
    uint32 ix = (uint32&)x;
    uint32 exp = (ix >> 23) & 0xFF;
    int32 log2 = int32(exp) - 127;

    return log2;
}

这比前面的例子快得多，但输出是无符号类型.

which is much faster than the previous example, but the output is unsigned type.

是否可以让这个函数返回一个double类型?

Is it possible to make this function return a double type?

提前致谢！

推荐答案

如果你只需要对数的整数部分，那么你可以直接从浮点数中提取出来.

If you just need the integer part of the logarithm, then you can extract that directly from the floating point number.

便携:

#include <cmath>

int log2_fast(double d) {
    int result;
    std::frexp(d, &result);
    return result-1;
}

可能更快，但依赖于未指定和未定义的行为:

Possibly faster, but relying on unspecified and undefined behaviour:

int log2_evil(double d) {
    return ((reinterpret_cast<unsigned long long&>(d) >> 52) & 0x7ff) - 1023;
}

这篇关于快速 log2(float x) 实现 C++的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！