我一直在研究。net反汇编和GCC源代码,但似乎在任何地方都找不到sin()和其他数学函数的实际实现…他们似乎总是在引用别的东西。

有人能帮我找到他们吗?我觉得不太可能运行C语言的所有硬件都支持三角函数,所以一定有软件算法,对吧?


我知道有几种方法可以计算函数,并且为了好玩,我写了自己的例程来使用泰勒级数计算函数。我很好奇生产语言是如何做到这一点的,因为我的所有实现总是慢几个数量级,尽管我认为我的算法非常聪明(显然不是)。


当前回答

计算正弦/余弦/正切其实很容易通过代码使用泰勒级数来实现。自己写一个只需5秒钟。

整个过程可以用这个方程来概括:

下面是我为C语言写的一些例程:

double _pow(double a, double b) {
    double c = 1;
    for (int i=0; i<b; i++)
        c *= a;
    return c;
}

double _fact(double x) {
    double ret = 1;
    for (int i=1; i<=x; i++) 
        ret *= i;
    return ret;
}

double _sin(double x) {
    double y = x;
    double s = -1;
    for (int i=3; i<=100; i+=2) {
        y+=s*(_pow(x,i)/_fact(i));
        s *= -1;
    }  
    return y;
}
double _cos(double x) {
    double y = 1;
    double s = -1;
    for (int i=2; i<=100; i+=2) {
        y+=s*(_pow(x,i)/_fact(i));
        s *= -1;
    }  
    return y;
}
double _tan(double x) {
     return (_sin(x)/_cos(x));  
}

其他回答

在GNU libm中,sin的实现依赖于系统。因此,您可以在sysdeps的适当子目录中找到每个平台的实现。

一个目录包含一个由IBM贡献的C语言实现。自2011年10月以来,这是在典型的x86-64 Linux系统上调用sin()时实际运行的代码。它显然比汇编指令中的f_f快。源代码:sysdeps/ieee754/dbl-64/s_sin.c,查找__sin (double x)。

这段代码非常复杂。没有一种软件算法在整个x值范围内尽可能快且准确,因此库实现了几种不同的算法,它的第一项工作是查看x并决定使用哪种算法。

When x is very very close to 0, sin(x) == x is the right answer. A bit further out, sin(x) uses the familiar Taylor series. However, this is only accurate near 0, so... When the angle is more than about 7°, a different algorithm is used, computing Taylor-series approximations for both sin(x) and cos(x), then using values from a precomputed table to refine the approximation. When |x| > 2, none of the above algorithms would work, so the code starts by computing some value closer to 0 that can be fed to sin or cos instead. There's yet another branch to deal with x being a NaN or infinity.

这段代码使用了一些我以前从未见过的数值技巧,尽管据我所知,它们可能在浮点专家中很有名。有时几行代码需要几段文字来解释。例如,这两条线

double t = (x * hpinv + toint);
double xn = t - toint;

(有时)用于将x减小到接近0的值,该值与x相差π/2的倍数,特别是xn × π/2。这种没有划分或分支的方式相当聪明。但是没有任何评论!


旧的32位版本的GCC/glibc使用fsin指令,这对于某些输入是非常不准确的。有一篇精彩的博客文章用两行代码说明了这一点。

fdlibm在纯C中实现sin要比glibc简单得多,而且注释很好。源代码:fdlibm/s_sin.c和fdlibm/k_sin.c

对于罪恶,用泰勒展开可以得到

Sin (x) = x - x^3/3!+ x ^ 5/5 !- x ^ 7/7 !+……(1)

您将继续添加项,直到它们之间的差异低于可接受的容忍水平,或者只是有限的步数(更快,但不太精确)。举个例子:

float sin(float x)
{
  float res=0, pow=x, fact=1;
  for(int i=0; i<5; ++i)
  {
    res+=pow/fact;
    pow*=-1*x*x;
    fact*=(2*(i+1))*(2*(i+1)+1);
  }

  return res;
}

注:(1)适用于小角度的近似值sin(x)=x。对于更大的角度,你需要计算越来越多的项才能得到可接受的结果。 你可以使用while参数并继续,以达到一定的准确性:

double sin (double x){
    int i = 1;
    double cur = x;
    double acc = 1;
    double fact= 1;
    double pow = x;
    while (fabs(acc) > .00000001 &&   i < 100){
        fact *= ((2*i)*(2*i+1));
        pow *= -1 * x*x; 
        acc =  pow / fact;
        cur += acc;
        i++;
    }
    return cur;

}

库函数的实际实现取决于特定的编译器和/或库提供程序。不管它是用硬件还是软件,不管它是不是泰勒展开,等等,都会有所不同。

我意识到这完全没有帮助。

如果你想要一个软件实现,而不是硬件实现,可以在《数值公式》的第5章中找到这个问题的明确答案。我的副本在一个盒子里,所以我不能给出细节,但简短的版本(如果我没记错的话)是你把tan(theta/2)作为你的基本操作,然后从那里计算其他的。计算是用级数近似完成的,但它比泰勒级数收敛得快得多。

抱歉,我没拿到书就想不起来了。

OK kiddies, time for the pros.... This is one of my biggest complaints with inexperienced software engineers. They come in calculating transcendental functions from scratch (using Taylor's series) as if nobody had ever done these calculations before in their lives. Not true. This is a well defined problem and has been approached thousands of times by very clever software and hardware engineers and has a well defined solution. Basically, most of the transcendental functions use Chebyshev Polynomials to calculate them. As to which polynomials are used depends on the circumstances. First, the bible on this matter is a book called "Computer Approximations" by Hart and Cheney. In that book, you can decide if you have a hardware adder, multiplier, divider, etc, and decide which operations are fastest. e.g. If you had a really fast divider, the fastest way to calculate sine might be P1(x)/P2(x) where P1, P2 are Chebyshev polynomials. Without the fast divider, it might be just P(x), where P has much more terms than P1 or P2....so it'd be slower. So, first step is to determine your hardware and what it can do. Then you choose the appropriate combination of Chebyshev polynomials (is usually of the form cos(ax) = aP(x) for cosine for example, again where P is a Chebyshev polynomial). Then you decide what decimal precision you want. e.g. if you want 7 digits precision, you look that up in the appropriate table in the book I mentioned, and it will give you (for precision = 7.33) a number N = 4 and a polynomial number 3502. N is the order of the polynomial (so it's p4.x^4 + p3.x^3 + p2.x^2 + p1.x + p0), because N=4. Then you look up the actual value of the p4,p3,p2,p1,p0 values in the back of the book under 3502 (they'll be in floating point). Then you implement your algorithm in software in the form: (((p4.x + p3).x + p2).x + p1).x + p0 ....and this is how you'd calculate cosine to 7 decimal places on that hardware.

请注意,在FPU中大多数硬件实现的超越操作通常涉及一些微码和类似的操作(取决于硬件)。 切比雪夫多项式用于大多数先验多项式,但不是全部。例:使用Newton raphson方法的两次迭代,首先使用查询表,使用平方根更快。 同样,《计算机逼近》这本书会告诉你。

If you plan on implmementing these functions, I'd recommend to anyone that they get a copy of that book. It really is the bible for these kinds of algorithms. Note that there are bunches of alternative means for calculating these values like cordics, etc, but these tend to be best for specific algorithms where you only need low precision. To guarantee the precision every time, the chebyshev polynomials are the way to go. Like I said, well defined problem. Has been solved for 50 years now.....and thats how it's done.

Now, that being said, there are techniques whereby the Chebyshev polynomials can be used to get a single precision result with a low degree polynomial (like the example for cosine above). Then, there are other techniques to interpolate between values to increase the accuracy without having to go to a much larger polynomial, such as "Gal's Accurate Tables Method". This latter technique is what the post referring to the ACM literature is referring to. But ultimately, the Chebyshev Polynomials are what are used to get 90% of the way there.

享受。