c# - 浮点运算稳定吗？

Question

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Answer 1

But what if the equation used to calculate the number is the same? can I assume the outcome would be the same too?

不，不一定。

特别是，在某些情况下，允许 JIT 使用更准确的中间表示 - 例如当原始数据为 64 位时为 80 位 - 而在其他情况下则不会。当以下任一情况为真时，这可能会导致看到不同的结果:

• 您的代码略有不同，例如使用局部变量而不是字段，这可以更改值是否存储在寄存器中。 (这是一个相对明显的例子；还有其他更微妙的例子可以影响事情，例如方法中存在 try block ......)
• 您在不同的处理器上执行(我曾经观察过 AMD 和 Intel 之间的差异；同一制造商的不同 CPU 之间也可能存在差异)
• 您正在执行不同的优化级别(例如，是否在调试器下)

来自 C# 5 规范第 4.1.6 节:

Floating-point operations may be performed with higher precision than the result type of the operation. For example, some hardware architectures support an "extended" or "long double" floating-point type with greater range and precision than the double type, and implicitly perform all floating-point operations using this higher precision type. Only at excessive cost in performance can such hardware architectures be made to perform floating-point operations with less precision, and rather than require an implementation to forfeit both performance and precision, C# allows a higher precision type to be used for all floating-point operations. Other than delivering more precise results, this rarely has any measurable effects. However, in expressions of the form x * y / z, where the multiplication produces a result that is outside the double range, but the subsequent division brings the temporary result back into the double range, the fact that the expression is evaluated in a higher range format may cause a finite result to be produced instead of an infinity.