请考虑 std :: exp 在标题 cmath /w/cpp/numeric"rel =" nofollow noreferrer>数字库.现在，请考虑C ++标准库的实现，例如 libstdc ++ .
Please consider std::exp defined in the header cmath in C++ numerics library. Now, please consider an implementation of the C++ Standard Library, say libstdc++.
考虑有各种算法来计算基本函数，例如算术几何平均迭代算法(用于计算指数函数)和其他三个算法，此处；
Considering there are various algorithms to compute the elementary functions, such as arithmetic-geometric mean iteration algorithm to compute the exponential function and three others shown here;
是否可以在 libstdc ++ 中命名用于计算指数函数的特定算法?
Could you please name the particular algorithm being used to compute the exponential function in libstdc++, if possible?
PS:恐怕我无法查明包含std :: exp实现的正确tarball或理解相关文件的内容.
PS: I could not pinpoint either the correct tarballs containing the std::exp implementation or comprehend the relevant file contents, I'm afraid.
推荐答案它根本不使用任何复杂的算法.请注意，std::exp仅针对非常有限的几种类型定义:float，double和long double +可转换为double的任何整数类型.这样就不必实施复杂的数学.
It doesn't use any intricate algorithm at all. Note that std::exp is only defined for a very limited number of types: float, double and long double + any Integral type that is castable to double. That makes it not necessary to implement complicated maths.
当前，它使用内置的__builtin_expf，可以从源代码.这将编译为对我计算机上的expf的调用，这是对来自glibc的libm的调用.让我们看看我们在其源代码中找到的内容.当我们搜索expf时，我们发现它在内部调用__ieee754_expf，这是与系统有关的实现. i686和x86_64都只包含一个glibc/sysdeps/ieee754/flt-32/e_expf.c，最终为我们提供了一个实现(为简洁起见，外观插入源文件
Currently, it uses the builtin __builtin_expf as can be verified from the source code. This compiles to a call to expf on my machine which is a call into libm coming from glibc. Let's see what we find in their source code. When we search for expf we find that this internally calls __ieee754_expf which is a system-dependant implementation. Both i686 and x86_64 just include a glibc/sysdeps/ieee754/flt-32/e_expf.c which finally gives us an implementation (reduced for brevity, the look into the sources
基本上是浮点数的3阶多项式近似值:
It is basically a order 3 polynomial approximation for floats:
static inline uint32_t top12 (float x) { return asuint (x) >> 20; } float __expf (float x) { uint64_t ki, t; /* double_t for better performance on targets with FLT_EVAL_METHOD==2. */ double_t kd, xd, z, r, r2, y, s; xd = (double_t) x; // [...] skipping fast under/overflow handling /* x*N/Ln2 = k + r with r in [-1/2, 1/2] and int k. */ z = InvLn2N * xd; /* Round and convert z to int, the result is in [-150*N, 128*N] and ideally ties-to-even rule is used, otherwise the magnitude of r can be bigger which gives larger approximation error. */ kd = roundtoint (z); ki = converttoint (z); r = z - kd; /* exp(x) = 2^(k/N) * 2^(r/N) ~= s * (C0*r^3 + C1*r^2 + C2*r + 1) */ t = T[ki % N]; t += ki << (52 - EXP2F_TABLE_BITS); s = asdouble (t); z = C[0] * r + C[1]; r2 = r * r; y = C[2] * r + 1; y = z * r2 + y; y = y * s; return (float) y; }类似地，对于128位long double，它是 7阶近似，对于double，他们使用更复杂的算法，我现在无法理解.
Similarly, for 128-bit long double, it's an order 7 approximation and for double they use more complicated algorithm that I can't make sense of right now.
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