在没有 Round Java 的情况下从 double 转换为 long

Convert from double to long without Round Java

提问by user2661808

I'm trying to convert a random double to a long without rounding or truncating it. First I change the double to a string, to know how many decimal places there are and then I change the value to long. My problem is that the number of the last decimal place is not always correct and I don't know the reason and how I can change it. Here is the code:

我正在尝试将随机双精度数转换为长整数而不舍入或截断它。首先，我将 double 更改为字符串，以了解有多少个小数位，然后将值更改为 long。我的问题是最后一个小数位的数字并不总是正确的，我不知道原因以及如何更改它。这是代码：

            double z=(double) myRandom(1, 20);
    long test;
    String s = Double.toString(z);
    test=(long) (Math.pow(10, s.length()-s.indexOf(".")-1)*z);
    System.out.println("z: "+z);    
    System.out.println("double converted to long: "+test);

And that's the output:

这就是输出：

d: 19.625014811604743

电话：19.625014811604743

double converted to long: 19625014811604744

double 转换为 long：19625014811604744

d: 9.143326452202839

d: 9.143326452202839

double converted to long: 9143326452202838

double 转换为 long：9143326452202838

d: 5.8964228376511105

d：5.8964228376511105

double converted to long: 58964228376511104

double 转换为 long：58964228376511104

d: 15.045936360299917

d: 15.045936360299917

double converted to long: 15045936360299918

double 转换为 long：15045936360299918

d: 14.147950026532694

d: 14.147950026532694

double converted to long: 14147950026532694

double 转换为 long：14147950026532694

回答by Andrei Juan

If your goal is to obtain a random long number, then I would strongly recommend using Random.nextLong() instead.

如果您的目标是获得一个随机长数，那么我强烈建议您改用 Random.nextLong()。

回答by gr3co

Editing your current implementation:

编辑您当前的实现：

double z=(double) myRandom(1, 20);
String s = Double.toString(z); //convert to string (edited thanks to comments)
s = s.replaceAll(".", ""); //remove the period
long test = Long.parseLong(s); //convert to long

Or if you just want a random LongI would recommend using Random.nextLong()instead.

或者，如果您只想要一个随机数，Long我建议您Random.nextLong()改用。

回答by Grammin

test=(long) (Math.pow(10, s.length()-s.indexOf(".")-1)*z);

Is messing up your double in the arithmetic, try something like this:

在算术中搞砸了你的双倍，尝试这样的事情：

public long convertDoubleToLong(double d){
  String str = Double.toString(d);
  return Long.parseLong(str.replace(".", ""));
}

回答by J.P. Armstrong

You can do something like this...

你可以做这样的事情......

double x = 19.625014811604743;
while(x % 1 != 0) x *= 10;
long y = (long) x;
System.out.println("x: "+ x + " y: "+ y);

回答by yshavit

The other answers have pointed out how you can fix this, but your other question was whyit happens. The answer to that is floating-point error.

其他答案指出了如何解决此问题，但您的另一个问题是为什么会发生这种情况。答案是浮点错误。

Floating-point numbers consist of three main parts: a sign (+ or -), a mantissa and an exponent. Essentially, the number is represented as essentially (sign) (mantissa) * 2^(exponent)(but not exactly -- see links below). As you can imagine, not all numbers fit snugly into this representation; some have to be approximated. The quintessential example is one-tenth, which in binary has an infinite repeating sequence and so has to be approximated in doubles.

浮点数由三个主要部分组成：符号（+ 或 -）、尾数和指数。从本质上讲，该数字表示为本质上(sign) (mantissa) * 2^(exponent)（但不完全是 - 请参阅下面的链接）。您可以想象，并非所有数字都适合这种表示方式。有些必须近似。典型的例子是十分之一，它在二进制中具有无限重复序列，因此必须以双精度近似。

At high values -- those greater than the mantissa can represnt -- there are some integers that can't be represented exactly. This happens at 2^53; any number greater than this has to be approximated, and the numbers you mention are such examples.

在高值时——那些大于尾数可以表示的值——有一些整数无法准确表示。这发生在 2^53；任何大于这个的数字都必须近似，你提到的数字就是这样的例子。

You don't even need Math.powto demonstrate this; literals will do just fine.

你甚至不需要Math.pow证明这一点；文字会做得很好。

public static void main(String[] args) {
  double d1 = 19.625014811604743;
  double d2 = 19625014811604743.0;

  System.out.println(d1);
  System.out.println(d2);
}

This will print:

这将打印：

19.625014811604743
1.9625014811604744E16

Notice the 4 at the end of that second value. 19625014811604744 is the closest number to 19625014811604743 that can be represented as a double.

注意第二个值末尾的 4。19625014811604744 是与 19625014811604743 最接近的数字，可以表示为双精度数。

Related reading:

相关阅读：

• http://floating-point-gui.de/Good introduction
• http://en.wikipedia.org/wiki/Binary64A bit more detail, about doubles specifically
• http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.htmlA bit dense, but important once you get the hang of things
• http://en.wikipedia.org/wiki/Floating_point

• http://floating-point-gui.de/好介绍
• http://en.wikipedia.org/wiki/Binary64更详细一点，特别是关于双打
• http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html有点密集，但一旦掌握了窍门就很重要
• http://en.wikipedia.org/wiki/Floating_point