如何在C＃中获得正确的除法结果？ [英] How to get the right division result in C# ?

问题描述

Hello Everyone，



我有以下问题：

Hello Everyone,

I have the following issue:

float increment = 0.00025f;
float sampleFreq = 1.f / increment;
[+] sampleFreq = 3999.99976

而不是4000为什么？



如何获得4000预期结果（不是通过舍入方法）？



当然是傻瓜。



非常感谢您提前。

MiQi



我尝试了什么：



我尝试过双倍的，相同的结果。

and not 4000 why ?

How may I have the 4000 expected result (not via a rounding method) ?

Certainly something stupid.

Thank you very much in advance.
MiQi

What I have tried:

I have tried with double, same result.

推荐答案

它有点数学，但浮点数和双精度数是浮点数类型，它们具有基数2指数，而不是我们人类习惯的基数10。有些数字不能准确地表示在这里（在我们的基数10中也是如此，想想1/3 = 0.33333333333 ......）并且在那里你会得到一些准确性损失。



这是使用浮点数的典型代表，你必须接受一个误差范围，并注意平等运算符之类的东西。



您可以尝试在.NET中使用十进制类型。这使用了基数为10的指数，因此其行为方式也是如此（大多数情况下），但它在占用空间和性能方面都有成本。

It's a bit mathematical, but floats and doubles are floating point types which have a base 2 exponent as opposed to the base 10 we humans are used to. Some numbers can't be represented exactly in this (and also in our base 10 for that matter, think 1/3 = 0.33333333333....) and where that happens you get some accuracy loss.

This is typical of working with floating point numbers, you have to accept a margin of error and be careful with things like equality operators.

You could try using the decimal type in .NET. This uses a base 10 exponent so behaves the way you would expect it too (mostly), but it comes with a cost in terms of footprint and performance.

因为你正在使用浮点数数据类型。它不能完全代表浮点数。阅读：这里 [ ^ ]和这里 [ ^ ]。



想一想。你如何用有限的位表示无限数量的值？你不能。牺牲必须使数据类型能够代表最常用的各种值，并提供一定程度的准确性。



你可以使用十进制类型 [ ^ ]。它是一个128位类型，为财务目的提供更高的准确性，但它可以表示的值范围更小。

Because you're using the float data type. It can't represent floating point numbers exactly. Read up on it: here[^] and here[^] .

Think about it. How you do represent an infinite number of values in a finite number of bits? You can't. Sacrifices have to be made to make a data type that can represent a wide range of values that are most commonly used and offer some degree of "accuracy".

You can use the decimal type[^] instead. It's a 128-bit type, made for higher accuracy for things like financial purposes, but has a smaller range of values it can represent.

float和double的精度是不够的。

The precision of float and double is not sufficient.

decimal increment = 0.00025M;
decimal sampleFreq = 1M / increment;

使用小数代替。

Use decimal instead.

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