在C＃中比较双重价值 [英] Comparing double values in C#

问题描述

我使用C＃。

我有一个名为x的双变量。 在code，X被赋予0.1的值，我检查它的'如果'的语句比较x和0.1

I've a double variable called x. In the code, x gets assigned a value of 0.1 and I check it in an 'if' statement comparing x and 0.1

if(x==0.1)
{
----
}

不幸的是它不进入if语句

Unfortunately it does not enter the if statement

1）我应该使用双或双???

1) Should I use Double or double???

2）什么是这背后的原因是什么？您能否提供一个解决方案吗？

2) What's the reason behind this? Can you suggest a solution for this?

推荐答案

这是一个标准的问题，由于计算机如何存储浮点值。搜索这里浮点问题，你会发现大量信息。

It's a standard problem due to how the computer stores floating point values. Search here for "floating point problem" and you'll find tons of information.

在短期 - 一个浮点/双精度不能储存0.1 precisely。这将永远是一个有点过。

In short - a float/double can't store 0.1 precisely. It will always be a little off.

您可以尝试使用十进制键入存储号码十进制格式。因此，0.1将重新presentable precisely。

You can try using the decimal type which stores numbers in decimal notation. Thus 0.1 will be representable precisely.

---

您想知道的原因：

浮点/双精度都为二进制小数，不是小数存储。为了说明：

Float/double are stored as binary fractions, not decimal fractions. To illustrate:

在十进制形式12.34（我们使用的）是指1 * 10 ¹ + 2 * 10 ⁰ + 3 * 10 ¹ + 4 * 10 ^-2 。计算机存储浮点数以相同的方式，除了使用基材2：10.01装置1 * 2 ¹ + 0 * 2 ⁰ + 0 * 2 ^{- 1} + 1 * 2 ^-2

12.34 in decimal notation (what we use) means 1*10¹+2*10⁰+3*10^-1+4*10^-2. The computer stores floating point numbers in the same way, except it uses base 2: 10.01 means 1*2¹+0*2⁰+0*2^-1+1*2^-2

现在，你可能知道，有一些不能被重新$ P $的一些数字与我们的十进制形式psented充分。例如利用十进制表示的三分之一是0.3333333 ...同样的事情发生在二进制表示法不同之处在于，不能pcisely重新presented $ P $的数字是不同的。其中之一是数量的1/10。在二进制表示法是0.000110011001100 ...

Now, you probably know that there are some numbers that cannot be represented fully with our decimal notation. For example, 1/3 in decimal notation is 0.3333333... The same thing happens in binary notation, except that the numbers that cannot be represented precisely are different. Among them is the number 1/10. In binary notation that is 0.000110011001100...

由于二进制符号不能存储它precisely，它被存储在一个倒圆的方式。因此，你的问题。

Since the binary notation cannot store it precisely, it is stored in a rounded-off way. Hence your problem.

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