为什么会有所不同在C＃中的浮点精度时圆括号分离时声明分开吗？ [英] Why differs floating-point precision in C# when separated by parantheses and when separated by statements?

问题描述

我知道如何浮点精度在正常的情况下工作，但我无意中发现在我的C＃代码一个奇怪的局面。

I am aware of how floating point precision works in the regular cases, but I stumbled on an odd situation in my C# code.

为什么不RESULT1和结果2完全相同的浮点值，这里

Why aren't result1 and result2 the exact same floating point value here?

const float A;   // Arbitrary value
const float B;   // Arbitrary value

float result1 = (A*B)*dt;

float result2 = (A*B); 
result2 *= dt;

从的此页面我想通浮动算法是左关联的，这意味着价值进行评估，并在左到右的方式计算。

From this page I figured float arithmetic was left-associative and that this means values are evaluated and calculated in a left-to-right manner.

完整的源代码涉及到XNA的四元数。我不认为这是我的相关常量是什么，什么VectorHelper.AddPitchRollYaw（）一样。如果我计算以同样的方式三角洲俯仰/卷/偏航角，但作为代码如下它没有通过测试通过就好了：

The full source code involves XNA's Quaternions. I don't think it's relevant what my constants are and what the VectorHelper.AddPitchRollYaw() does. The test passes just fine if I calculate the delta pitch/roll/yaw angles in the same manner, but as the code is below it does not pass:

X
  Expected: 0.275153548f
  But was:  0.275153786f

[TestFixture]
    internal class QuaternionPrecisionTest
    {
        [Test]
        public void Test()
        {
            JoystickInput input;
            input.Pitch = 0.312312432f;
            input.Roll = 0.512312432f;
            input.Yaw = 0.912312432f;
            const float dt = 0.017001f;

            float pitchRate = input.Pitch * PhysicsConstants.MaxPitchRate;
            float rollRate = input.Roll * PhysicsConstants.MaxRollRate;
            float yawRate = input.Yaw * PhysicsConstants.MaxYawRate;

            Quaternion orient1 = Quaternion.Identity;
            Quaternion orient2 = Quaternion.Identity;

            for (int i = 0; i < 10000; i++)
            {
                float deltaPitch = 
                      (input.Pitch * PhysicsConstants.MaxPitchRate) * dt;
                float deltaRoll = 
                      (input.Roll * PhysicsConstants.MaxRollRate) * dt;
                float deltaYaw = 
                      (input.Yaw * PhysicsConstants.MaxYawRate) * dt;

                // Add deltas of pitch, roll and yaw to the rotation matrix
                orient1 = VectorHelper.AddPitchRollYaw(
                                orient1, deltaPitch, deltaRoll, deltaYaw);

                deltaPitch = pitchRate * dt;
                deltaRoll = rollRate * dt;
                deltaYaw = yawRate * dt;
                orient2 = VectorHelper.AddPitchRollYaw(
                                orient2, deltaPitch, deltaRoll, deltaYaw);
            }

            Assert.AreEqual(orient1.X, orient2.X, "X");
            Assert.AreEqual(orient1.Y, orient2.Y, "Y");
            Assert.AreEqual(orient1.Z, orient2.Z, "Z");
            Assert.AreEqual(orient1.W, orient2.W, "W");
        }
    }

当然，误差小，仅在提出自己大量的迭代，但它给我带来一些伟大的headackes。

Granted, the error is small and only presents itself after a large number of iterations, but it has caused me some great headackes.

推荐答案

我找不到一个参考支持这一了，但我认为这是由于以下几点：

I couldn't find a reference to back this up but I think it is due to the following:

• 浮点运算在硬件提供的精确计算，这意味着他们可以用比浮动的更高的精度进行。


• 分配到中间 RESULT2 变量力量四舍五入回到浮动的精度，但对于单个表达式 rsult1 完全计算的本地精密被舍去了。

• float operations are calculated in the precision available in the hardware, that means they can be done with a greater precision than that of float.
• the assignment to the intermediate result2 variable forces rounding back to float precision, but the single expression for rsult1 is computed entirely in native precision before being rounded down.

在一个侧面说明，测试float或double与 == 总是危险的。微软单元测试提供了上午 Assert.AreEqual（浮点预期，实际浮动，浮动三角洲）在这里，你会用适当的增量解决这个问题（基于float.Epsilon）

On a side note, testing float or double with == is always dangerous. The Microsoft Unit testing provides for am Assert.AreEqual(float expected, float actual,float delta) where you would solve this problem with a suitable delta (based on float.Epsilon).

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