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不使用浮点型在C中建立对数函数

[英]Building a logarithm function in C without using float type

I need to rewrite the log function (base 2 or base 10 doesn't matter which) without using float type, but I need to get the precision of few decimal digits after the decimal point. 我需要在不使用float类型的情况下重写日志函数(以2基数还是以10基数无关紧要),但是我需要获得小数点后几位小数位数的精度。 ( like a float * 100 to get 2 decimals inside integer type eg: if the 1.4352 would be the result, my function should return something like 143 ( int type) and I know that the last 2 numbers are the decimals. (例如float * 100以在整数类型内获取2 1.4352数,例如:如果结果为1.4352 ,则我的函数应返回类似143int类型)的值,并且我知道最后2个数字是小数。

I found over the stackoverflow some methods like: 我在stackoverflow上发现了一些方法,例如:

but all of them returns int precision ( avoiding the decimals ). 但它们都返回int precision(避免使用小数点)。

I have no idea how to approach this so the question is: 我不知道如何解决这个问题,所以问题是:

How to encode (and or change) integer log implementation to support decimal result? 如何编码(和/或更改)整数log实现以支持十进制结果?

You need to use fixed point precision/arithmetics/math for this. 为此,您需要使用定点精度/算术/数学。 It means you use integer type variables but some of the bits are after the decimal point. 这意味着您使用整数类型变量,但某些位在小数点后。

for example let assume 8 decimal bits so operations are done like this: 例如,假设8个十进制位,则操作如下所示:

a = number1*256
b = number2*256
c=a+b // +
c=a-b // -
c=(a*b)>>8 // *
c=(a/b)<<8 // /

Here simple fixed point log2 example via binary search in C++ : 这是在C ++中通过二进制搜索的简单定点log2示例:

//---------------------------------------------------------------------------
const DWORD _fx32_bits      =32;                            // all bits count
const DWORD _fx32_fract_bits= 8;                            // fractional bits count
const DWORD _fx32_integ_bits=_fx32_bits-_fx32_fract_bits;   // integer bits count
//---------------------------------------------------------------------------
const DWORD _fx32_one       =1<<_fx32_fract_bits;           // constant=1.0 (fixed point)
const DWORD _fx32_fract_mask=_fx32_one-1;                   // fractional bits mask
const DWORD _fx32_integ_mask=0xFFFFFFFF-_fx32_fract_mask;   // integer bits mask
const DWORD _fx32_MSB_mask=1<<(_fx32_bits-1);               // max unsigned bit mask
//---------------------------------------------------------------------------
DWORD bits(DWORD p)             // count how many bits is p
    {
    DWORD m=0x80000000; DWORD b=32;
    for (;m;m>>=1,b--)
     if (p>=m) break;
    return b;
    }
//---------------------------------------------------------------------------
DWORD fx32_mul(DWORD x,DWORD y)
    {
    // this should be done in asm with 64 bit result !!!
    DWORD a=x,b=y;              // asm has access only to local variables
    asm {                       // compute (a*b)>>_fx32_fract
        mov eax,a               // eax=a
        mov ebx,b               // ebx=b
        mul eax,ebx             // (edx,eax)=eax*ebx
        mov ebx,_fx32_one
        div ebx                 // eax=(edx,eax)>>_fx32_fract
        mov a,eax;
        }
    return a;
    // you can also do this instead but unless done on 64bit variable will overflow
    return (x*y)>>_fx32_fract_bits;
    }
//---------------------------------------------------------------------------
DWORD fx32_sqrt(const DWORD &x) // unsigned fixed point sqrt
    {
    DWORD m,a;
    if (!x) return 0;
    m=bits(x);                  // integer bits
    if (m>_fx32_fract_bits) m-=_fx32_fract_bits; else m=0;
    m>>=1;                      // sqrt integer result is half of x integer bits
    m=_fx32_one<<m;             // MSB of result mask
    for (a=0;m;m>>=1)           // test bits from MSB to 0
        {
        a|=m;                   // bit set
        if (fx32_mul(a,a)>x)    // if result is too big
         a^=m;                  // bit clear
        }
    return a;
    }
//---------------------------------------------------------------------------
DWORD fx32_exp2(DWORD y)       // 2^y
    {
    // handle special cases
    if (!y) return _fx32_one;                    // 2^0 = 1
    if (y==_fx32_one) return 2;                  // 2^1 = 2
    DWORD m,a,b,_y;
    // handle the signs
    _y=y&_fx32_fract_mask;      // _y fractional part of exponent
     y=y&_fx32_integ_mask;      //  y integer part of exponent
    a=_fx32_one;                // ini result
    // powering by squaring x^y
    if (y)
        {
        for (m=_fx32_MSB_mask;(m>_fx32_one)&&(m>y);m>>=1);     // find mask of highest bit of exponent
        for (;m>=_fx32_one;m>>=1)
            {
            a=fx32_mul(a,a);
            if (DWORD(y&m)) a<<=1;  // a*=2
            }
        }
    // powering by rooting x^_y
    if (_y)
        {
        for (b=2<<_fx32_fract_bits,m=_fx32_one>>1;m;m>>=1)      // use only fractional part
            {
            b=fx32_sqrt(b);
            if (DWORD(_y&m)) a=fx32_mul(a,b);
            }
        }
    return a;
    }
//---------------------------------------------------------------------------
DWORD fx32_log2(DWORD x)    // = log2(x)
    {
    DWORD y,m;
    // binary search from highest possible integer power of 2 to avoid overflows (log2(integer bits)-1)
    for (y=0,m=_fx32_one<<(bits(_fx32_integ_bits)-1);m;m>>=1)
        {
        y|=m;   // set bit
        if (fx32_exp2(y)>x) y^=m; // clear bit if result too big
        }
    return y;
    }
//---------------------------------------------------------------------------

Here simple test (using floats just for loading and printing you can handle booth on integers too, or by compiler evaluated constants): 这是一个简单的测试(使用浮点数仅用于加载和打印,您也可以处理整数值的展位,或者通过编译器评估的常数):

float(fx32_log2(float(125.67*float(_fx32_one)))) / float(_fx32_one)

This evaluates: log2(125.67) = 6.98828125 my win calc returns 6.97349648 which is pretty close. 结果为: log2(125.67) = 6.98828125我的获胜计算返回6.97349648 ,这非常接近。 More precise result you need more fractional bits you need to use. 更精确的结果是您需要使用更多的小数位。 Int and compile time evaluation float example: Int和编译时间评估浮点示例:

(100*fx32_log2(125.67*_fx32_one))>>_fx32_fract_bits

returns 698 which means 6.98 as we multiplied by 100 . 返回698表示6.98乘以100 You can also write your own load and print function to convert between fixed point and string directly. 您还可以编写自己的加载和打印功能,以在定点字符串之间直接转换。

To change precision just play with _fx32_fract_bits constant. 要更改精度,只需使用_fx32_fract_bits常数即可。 Anyway if your C++ does not know DWORD it is just 32bit unsigned int . 无论如何,如果您的C ++不知道DWORD ,则它只是32位unsigned int If you are using different type (like 16 or 64 bit) then just change the constants accordingly. 如果您使用其他类型(例如1664位),则只需相应地更改常量。

For more info take a look at: 有关更多信息,请查看:

[Edit2] fx32_mul on 32bit arithmetics without asm base 2^16 O(n^2) [Edit2] fx32_mul的32位算术,没有asm base 2^16 O(n^2)

DWORD fx32_mul(DWORD x,DWORD y)
    {
    const int _h=1; // this is MSW,LSW order platform dependent So swap 0,1 if your platform is different
    const int _l=0;
    union _u
        {
        DWORD u32;
        WORD u16[2];
        }u;
    DWORD al,ah,bl,bh;
    DWORD c0,c1,c2,c3;
    // separate 2^16 base digits
    u.u32=x; al=u.u16[_l]; ah=u.u16[_h];
    u.u32=y; bl=u.u16[_l]; bh=u.u16[_h];
    // multiplication (al+ah<<1)*(bl+bh<<1) = al*bl + al*bh<<1 + ah*bl<<1 + ah*bh<<2
    c0=(al*bl);
    c1=(al*bh)+(ah*bl);
    c2=(ah*bh);
    c3= 0;
    // propagate 2^16 overflows (backward to avoid overflow)
    c3+=c2>>16; c2&=0x0000FFFF;
    c2+=c1>>16; c1&=0x0000FFFF;
    c1+=c0>>16; c0&=0x0000FFFF;
    // propagate 2^16 overflows (normaly to recover from secondary overflow)
    c2+=c1>>16; c1&=0x0000FFFF;
    c3+=c2>>16; c2&=0x0000FFFF;
    // (c3,c2,c1,c0) >> _fx32_fract_bits
    u.u16[_l]=c0; u.u16[_h]=c1; c0=u.u32;
    u.u16[_l]=c2; u.u16[_h]=c3; c1=u.u32;
    c0 =(c0&_fx32_integ_mask)>>_fx32_fract_bits;
    c0|=(c1&_fx32_fract_mask)<<_fx32_integ_bits;
    return c0;
    }

In case you do not have WORD,DWORD add this to start of code 如果您没有WORD,DWORD将此添加到代码开头

typedef unsigned __int32 DWORD;
typedef unsigned __int16 WORD;

or this: 或这个:

typedef uint32_t DWORD;
typedef uint16_t WORD;

[Edit3] fx32_mul debug info [Edit3] fx32_mul调试信息

let call and trace/breakpoint this (15 fractional bits): 让我们调用并跟踪/断点(15个小数位):

fx32_mul(0x00123400,0x00230056);

Which is: 这是:

0x00123400/32768 * 0x00230056/32768 =
36 * 70.00262451171875 = 2520.094482421875

So: 所以:

DWORD fx32_mul(DWORD x,DWORD y) // x=0x00123400 y=0x00230056
    {
    const int _h=1;
    const int _l=0;
    union _u
        {
        DWORD u32;
        WORD u16[2];
        }u;
    DWORD al,ah,bl,bh;
    DWORD c0,c1,c2,c3;
    // separate 2^16 base digits
    u.u32=x; al=u.u16[_l]; ah=u.u16[_h]; // al=0x3400 ah=0x0012
    u.u32=y; bl=u.u16[_l]; bh=u.u16[_h]; // bl=0x0056 bh=0x0023
    // multiplication (al+ah<<1)*(bl+bh<<1) = al*bl + al*bh<<1 + ah*bl<<1 + ah*bh<<2
    c0=(al*bl);        // c0=0x00117800
    c1=(al*bh)+(ah*bl);// c1=0x0007220C
    c2=(ah*bh);        // c2=0x00000276
    c3= 0;             // c3=0x00000000
    // propagate 2^16 overflows (backward to avoid overflow)
    c3+=c2>>16; c2&=0x0000FFFF; // c3=0x00000000 c2=0x00000276
    c2+=c1>>16; c1&=0x0000FFFF; // c2=0x0000027D c1=0x0000220C
    c1+=c0>>16; c0&=0x0000FFFF; // c1=0x0000221D c0=0x00007800
    // propagate 2^16 overflows (normaly to recover from secondary overflow)
    c2+=c1>>16; c1&=0x0000FFFF; // c2=0x0000027D c1=0x0000221D
    c3+=c2>>16; c2&=0x0000FFFF; // c3=0x00000000 c2=0x0000027D
    // (c3,c2,c1,c0) >> _fx32_fract_bits
    u.u16[_l]=c0; u.u16[_h]=c1; c0=u.u32; // c0=0x221D7800
    u.u16[_l]=c2; u.u16[_h]=c3; c1=u.u32; // c1=0x0000027D
    c0 =(c0&_fx32_integ_mask)>>_fx32_fract_bits; // c0=0x0000443A
    c0|=(c1&_fx32_fract_mask)<<_fx32_integ_bits; // c0=0x04FA443A
    return c0; // 0x04FA443A -> 83510330/32768 = 2548.53302001953125
    }

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