将大数归一到小归一化数 [英] Bringing a big number to a small normalized number

问题描述

你好，我有这个代码:

Hello, i have this code:

private void button1_Click(object sender, EventArgs e)
        {

            double number =Convert.ToDouble( textBox1.Text);
            string s = (Math.Round(number, 3)*1000).ToString();

            int norm=1;
            for(int i=0;i<=s.Length-1;i++)
            {
                norm*=10;
            }
            label1.Text =Math.Round((Convert.ToDouble(s)/norm),3).ToString() ;

        }

这应该带来一个数字，例如11256,125698到1,125或18668951到1,866(理论上)等等.
理论作品(至少在我看来)并且数量很少，但是当我尝试18668951时，它给了我意想不到的效果(15.356).谁能告诉我我在哪里搞砸了? (加倍的数字不足以使用这么大的数字吗?)

this should bring a number like 11256,125698 to 1,125 or 18668951 to 1,866 (theorethicaly) and so on.
Theorethical works (at least for my mind) and with small numbers also but when i tried on 18668951 it gaved me an unexpected rezult(15.356). Can anyone tell me where i messed up? (double isnt enough to work with such big number?)

推荐答案

否，当我尝试过:

No, when I tried it:

double number = Convert.ToDouble("18668951");
string s = (Math.Round(number, 3) * 1000).ToString();

它给定为"18668951000"
因此，您尝试创建一个溢出的int(32位).
尝试将norm设置为Int64.

[edit]错字-OriginalGriff [/edit]

It gave s as "18668951000"
So you are trying to create an int (32 bits) which overflows.
Try making norm an Int64 instead.

[edit]Typo - OriginalGriff[/edit]

怎么了?

double l = Math.Ceiling(Math.Log10(number));
double n = number / Math.Pow(10,l-1);
label1.Text = Math.Round(n,3).ToString();

?

请注意(如果需要)，在特殊情况下，您必须处理十的幂(例如10, 100, 1000).

?

Note (if needed) you have to handle as special cases the powers of ten (e.g. 10, 100, 1000).

您好，
尝试使用此链接，您可以在其中找到许多解决问题的方法

如何在窗口应用程序中形成文本框值.... [ ^ ]

好吧，我建议您简化代码:

将double类型的变量类型用作"Convert.ToDouble(s)"的值，因为它在Math.Round()中太复杂了. 有时，此声明

Hello,
Try this link where you have many solutions for your problem

how to formate textbox value .....in window application[^]

Allso I sugest that you simplify code :

Use variable type of double for value of "Convert.ToDouble(s)" because it is too complex inside Math.Round()
Sometimes this statment

for(int i=0;i<=s.Length-1;i++)

无法正常工作(我仍然找不到原因)，所以我将

won''t work correct ( I still didn''t find out why ) so I use

for(int i=0;i<=s.Length-1;i=i+1)

在为变量范数选择int类型时出现错误，它应该为double类型，因为int类型对于在工作期间可能出现的数字范围而言太小程序.
避免在程序中使用的同一命令中进行变量声明，
示例:

And you have error in selecting type of int for variable norm, it should be type of double because int type is too small for a range of numbers that may appear during the work of program.
Avoid variable declaration in the same command that you use in the program,
exmpl :

for(int i=0;i<=s.Length-1;i++)

您所有人的状况都出现了错误
为了停止for..next循环，它应该是i< s.Lenght-1

我已经更改了您的代码，因此现在可以将示例18668951的结果转换为1,867，将11256,125698的结果转换为1,126，结果应该是这样.

看起来像这样:

You allso have error in condition
for stopping the for..next loop, it should be i < s.Lenght-1

I have changed your code so now it gives this result for your example 18668951 to 1,867 or 11256,125698 to 1,126 and it should be like that.

It looks like this :

void Button1Click(object sender, EventArgs e)
{
    //
    // Declaration of variables
    //
    string s = "";
    double norm = 0;
    double number = 0;
    double s1 = 0;
    int i = 0;
    int lenght = 0;

        s = textBox1.Text;
        number = Convert.ToDouble(s);
        s1 = Math.Round(number,3)*1000;
        lenght = s1.ToString().Length;
        norm=1;
        for(i=0;i<lenght-1;i=i+1)
        {
            norm = norm * 10;
        }
        label1.Text =Math.Round((s1/norm),3).ToString() ;
}

如果输入的数字根本不是数字，则应该注意输入的数字的相关性.Convert.ToDouble(s)将不起作用.






艾尔最好的
PerićŽeljko

You should allso take care of relevance of entered number , if it is not number at all Convert.ToDouble(s) won''t work.






Al the best
Perić Željko

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