c - 跳过 IEEE 754 中的 "test for zero"检查

Question

我在一本关于计算机图形学的流行书中读到过这个，

There are many numeric computations that become much simpler if the programmer takes advantage of the IEEE rules. For example, consider the expression:

a = 1 / (1/b + 1/c)

Such expressions arise with resistors and lenses. If divide-by-zero resulted in a program crash (as was true in many systems before IEEE floating-point), then two if statements would be required to check for small or zero values of b or c. Instead, with IEEE floating-point, if b or c is zero, we will get a zero value for a as desired.

但是 b=+0 和 c=-0 的情况呢？然后是a=1/inf-inf=nan。这本书关于这种优化是错误的，还是我误解了什么？看来我们仍然需要至少检查一次 b 和 c 的标志，而不是根本不检查。

编辑 评论中的一个建议是对 NaN 进行后检查。这是“利用”这些数字类型的惯用方式吗？

bool is_nan(float x) { return x != x; }

float optic_thing(float b, float c) {
  float result = 1.0f / (1.0f/b + 1.0f/c);
  if (is_nan(result)) result = 0;
  return result;
}

Answer 1

But what about the case where b=+0 and c=-0?

通过添加 0.0 将 -0.0 转换为 +0.0 而无需分支。

int main(void) {
  double b = +0.0;
  double c = -0.0;
  printf("%e\n", b);
  printf("%e\n", c);
  printf("%e\n", 1.0/(1.0/b + 1.0/c));
  b += 0.0;
  c += 0.0;
  printf("%e\n", 1.0/(1.0/b + 1.0/c));
  return 0;
}

输出

0.000000e+00
-0.000000e+00
nan
0.000000e+00

---

[编辑] 另一方面，任何接近 0.0 但不是 0.0 的值都可能是数字伪像而不是准确的电阻值.对于像 b = DBL_TRUE_MIN; 这样的 tiny 值对，上面的代码仍然存在问题。 c = -b; 问题是1.0/tiny --> Infinity and +Infinity + -Infinity --> 南。可以求助于使用更宽的 float 进行商计算或缩小操作数。

  b += 0.0;
  c += 0.0;
  printf("%Le\n", 1.0/(1.0L/b + 1.0L/c));

  // Lose some precision, but we are talking about real resistors/lenses here.
  b = (float) b + 0.0f;
  c = (float) c + 0.0f;
  printf("%e\n", 1.0/(1.0/b + 1.0/c));