是否有一个x的浮点值，其中x-x == 0是否为false？ [英] Is there a floating point value of x, for which x-x == 0 is false?

问题描述

在大多数情况下，我明白浮点比较测试应该使用超过一定范围的值（abs（x-y）ε）来实现，但是自减是否意味着结果将是零？

  //可以触发断言吗？ 
 float x = //？ 
 assert（xx == 0）
  

我的猜测是nan / inf可能是特殊的但是我对简单值发生了什么感兴趣。

编辑：

$ b $如果有人可以引用一个引用（IEEE浮点标准），我很乐意选择一个答案？

解决方案

正如你所暗示的， inf-inf 是 NaN ，它不等于零。同样， NaN - NaN 是 NaN 。但是，对于任何有限的浮点数 x ， x - x == 0.0 在舍入模式下， x - x 的结果可能是负的零，但是负的零比较等于 0.0 in浮点运算）。

编辑：给一个明确的标准引用有点棘手，因为这是规则的一个紧急属性在IEEE-754标准中阐述。具体来说，从第5条中定义的操作被正确舍入的要求出发。减法就是这样一个操作（第5.4.1节算术运算），正确舍入的结果 x - x 是适当符号的零（6.3节，当
和
两个操作数的差值（或者
）的两个操作数的总和为
$ b

与
完全相同的零），除了
roundTowardNegative之外，所有
舍入方向属性的总和（或
差值）的符号应为+0;在
属性下，一个确切的
总和（或差额）的符号应该是-0.

因此， x - x 的结果必须是 +/- 0 ，因此必须与 0.0 （5.11节第2段）：

lockquote

比较应忽略零符号。 / b>

进一步编辑：这并不是说错误的编译器不能引发该断言。你的问题是模棱两可的没有有限的浮点数 x ，所以 x - x == 0 为false。但是，这不是您发布的代码所检查的内容;它检查C样式语言中的某个表达式是否可以评估为非零值;特别是在某些平台上，在某些（不明智的）编译器优化的情况下，表达式中的变量 x 的两个实例可能有不同的 值，导致断言失败（特别是如果 x 是一些计算的结果，而不是一个常量，可表示的值）。这是这些平台上的数字模型中的一个错误，但这并不意味着它不会发生。

In most cases, I understand that a floating point comparison test should be implemented using over a range of values (abs(x-y) < epsilon), but does self subtraction imply that the result will be zero?

// can the assertion be triggered?
float x = //?;
assert( x-x == 0 )

My guess is that nan/inf might be special cases, but I'm more interested in what happens for simple values.

edit:

I'm happy to pick an answer if someone can cite a reference (IEEE floating point standard)?

解决方案

As you hinted, inf - inf is NaN, which is not equal to zero. Similarly, NaN - NaN is NaN. It is true, however, that for any finite floating-point number x, x - x == 0.0 (depending on the rounding mode, the result of x - x might be negative zero, but negative zero compares equal to 0.0 in floating-point arithmetic).

Edit: it's a little tricky to give a clear standards reference, because this is an emergent property of the rules set forth in the IEEE-754 standard. Specifically, it follows from the requirement that operations defined in Clause 5 be correctly rounded. Subtraction is such an operation (Section 5.4.1 "Arithmetic operations"), and the correctly-rounded result of x - x is a zero of the appropriate sign (Section 6.3, paragraph 3):

When the sum of two operands with opposite signs (or the difference of two operands with like signs) is exactly zero, the sign of that sum (or difference) shall be +0 in all rounding-direction attributes except roundTowardNegative; under that attribute, the sign of an exact zero sum (or difference) shall be −0.

So the result of x - x must be +/- 0, and therefore must compare equal to 0.0 (Section 5.11, paragraph 2):

Comparisons shall ignore the sign of zero.

Further Edit: That's not to say that a buggy compiler couldn't cause that assert to fire. Your question is ambiguous; there is no finite floating point number x such that x - x == 0 is false. However, that's not what the code that you posted checks; it checks whether or not a certain expression in a C-style language can evaluate to a non-zero value; in particular, on certain platforms, with certain (ill-conceived) compiler optimizations, the two instances of the variable x in that expression might have different values, causing the assertion to fail (especially if x is the result of some computation, instead of a constant, representable value). This is a bug in the numerics model on those platforms, but that doesn't mean that it can't happen.

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