否定浮点数总是安全的吗

Question

考虑：

double f = foo();
double g = -f;

其中 foo() 可以 return 分配给 f 的任何内容。

double g = -f; 在 C 和 C++ 中安全吗？对于 IEEE 754 类型，它显然是，但 C 和 C++ 不限制浮点实现（不像 Java）。

如果它是安全的，那么 -g 是否总是将 true 与 foo() 进行比较？

（以上对2的补码中的int不成立）

Answer 1

浮动类型由 C 标准（至少草案 N1570）的 §5.2.4.2.2 定义：

The characteristics of floating types are defined in terms of a model that describes a representation of floating-point numbers and values that provide information about an implementation’s floating-point arithmetic.21) The following parameters are used to define the model for each floating-point type:

s sign (±1)

b base or radix of exponent representation (an integer > 1)

e exponent (an integer between a minimum emin and a maximum emax)

p precision (the number of base-b digits in the significand)

fk nonnegative integers less than b (the significand digits)

A floating-point number (x) is defined by the following model:

所以是的，该表达式是有点安全，除非 foo 返回了 "other kinds of floating point numbers" 标准未规定：

In addition to normalized floating-point numbers ( f1 > 0 if x ≠ 0), floating types may be able to contain other kinds of floating-point numbers, such as subnormal floating-point numbers (x ≠ 0, e = emin, f1 = 0) and unnormalized floating-point numbers (x ≠ 0, e > emin, f1 = 0), and values that are not floating-point numbers, such as infinities and NaNs.

和：

An implementation may give zero and values that are not floating-point numbers (such as infinities and NaNs) a sign or may leave them unsigned.

对此可能还有其他注意事项，但标准对这些类型的特征进行了相当多的详细说明。您可以在 this publicly available draft.

中阅读有关它们的所有信息

否定浮点数总是安全的吗

Is it always safe to negate a floating point number

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