[英]How to use java.math.BigInteger in jjs / Nashorn?
I would like to use java.math.BigInteger in a nashorn / jss JavaScript. 我想在nashorn / jss JavaScript中使用java.math.BigInteger。
By way of example, let's say I want to calculate Fibonacci sequence numbers. 举例来说,假设我想计算Fibonacci序列号。 Numbers will need to remain exact, even if they become very large. 数字将需要保持精确,即使它们变得非常大。
Working Java code looks like this: 工作Java代码如下所示:
public static BigInteger fibonacci(int n) {
BigInteger prev = new BigInteger("0");
if (n == 0) return prev;
BigInteger next = new BigInteger("1");
if (n == 1) return next;
BigInteger fib = null;
int i;
for (i = 1; i < n; i++) {
fib = prev.add(next);
prev = next;
next = fib;
}
return fib;
}
We can test with: 我们可以测试:
So far so good. 到现在为止还挺好。
Equivalent JavaScript code below: 下面的等效JavaScript代码:
function fibonacci(n) {
var BigInteger = Java.type("java.math.BigInteger");
prev = new BigInteger("0");
if (n == 0) return prev;
next = new BigInteger("1");
if (n == 1) return next;
var i, fib = null;
for (i = 1; i < n; i++) {
fib = prev.add(next);
prev = next;
next = fib;
}
return fib;
}
Now we get: 现在我们得到:
Note that the value for 79 is one off - it's wrong. 请注意,79的值是一次性 - 这是错误的。
I suspect the problem is that somewhere, the BigNumber values are re-interpreted as plain JavaScript Numbers. 我怀疑问题是在某处,BigNumber值被重新解释为纯JavaScript编号。 (by "somewhere" I suspect this already happens as the supposedly BigInteger is passed to the .add method) (通过“某个地方”我怀疑这已经发生,因为所谓的BigInteger被传递给.add方法)
For example, if I you do: 例如,如果我这样做:
var BigInteger = Java.type("java.math.BigInteger");
print(new BigInteger("14472334024676221"));
The output is 14472334024676220
, not 14472334024676221
. 输出为14472334024676220
,而不是14472334024676221
。 This happens even if I explicitly call .toString()
on the BigInteger object. 即使我在BigInteger对象上显式调用.toString()
,也会发生这种情况。
How do I get past this? 我如何通过这个?
UPDATE: @Dici asked if I looked for a threshold. 更新:@Dici问我是否找了一个门槛。 I did - I found: 我做了 - 我发现:
var str, BigInteger = Java.type("java.math.BigInteger");
str = "9999999999999998";
print(str + ": " + new BigInteger(str));
str = "9999999999999999";
print(str + ": " + new BigInteger(str));
will output: 将输出:
I'm not sure it it's a matter of "treshold", or of some particular numbers having inaccuracies though. 我不确定它是“门槛”的问题,还是某些特定数字有不准确的问题。
UPDATE 2: 更新2:
This is now reported as a bug: https://bugs.openjdk.java.net/browse/JDK-8146264 Bug report was done by a Oracle JDK/Nashorn developer so I guess it's the real thing. 现在报告为一个错误: https : //bugs.openjdk.java.net/browse/JDK-8146264错误报告由Oracle JDK / Nashorn开发人员完成,所以我猜这是真的。 Keeping my fingers crossed. 我的手指交叉。
I took your example: 我举了你的例子:
var BigInteger = Java.type("java.math.BigInteger");
print(new BigInteger("14472334024676221"));
Started the program in debug mode and noticed that toString
method of BigInteger
was not used. 在调试模式下启动程序并注意到没有使用BigInteger
toString
方法。 So I created a simple class: 所以我创建了一个简单的类:
public class ToString {
private final BigInteger x;
public ToString(BigInteger x) {
this.x = x;
}
@Override
public String toString() {
return x.toString();
}
}
And used it in order to output the BigInteger
, and it worked: 并使用它来输出BigInteger
,它工作:
ScriptEngineManager scriptEngineManager = new ScriptEngineManager();
ScriptEngine jsEngine = scriptEngineManager.getEngineFactories().get(0).getScriptEngine();
String script = "var BigInteger = Java.type(\"java.math.BigInteger\");\n" +
"var ToString = Java.type(\"com.stackoverflow.inner.ToString\");\n" +
"var ts = new ToString(new BigInteger(\"14472334024676221\"));\n" +
"print(ts);";
jsEngine.eval(script); // prints 14472334024676221
Then I suspected that Nashorn used some intermediate conversion before converting BigInteger
to String
so I created a breakpoint at BigInteger.doubleValue()
and it triggered when bare BigInteger
was printed. 然后我怀疑Nashorn在将BigInteger
转换为String
之前使用了一些中间转换,所以我在BigInteger.doubleValue()
创建了一个断点,并在打印裸BigInteger
时触发。 Here is the problematic stack trace to let you understand Nashorn's logic: 这是有问题的堆栈跟踪,让您了解Nashorn的逻辑:
at java.math.BigInteger.doubleValue(BigInteger.java:3888)
at jdk.nashorn.internal.runtime.JSType.toStringImpl(JSType.java:976)
at jdk.nashorn.internal.runtime.JSType.toString(JSType.java:327)
at jdk.nashorn.internal.runtime.JSType.toCharSequence(JSType.java:341)
at jdk.nashorn.internal.objects.NativeString.constructor(NativeString.java:1140)
And the problematic Nashorn's code JSType.toStringImpl
: 有问题的Nashorn的代码JSType.toStringImpl
:
if (obj instanceof Number) {
return toString(((Number)obj).doubleValue());
}
Yes, this is an issue. 是的,这是一个问题。 A bug has been filed -> https://bugs.openjdk.java.net/browse/JDK-8146264 已经提交了一个错误 - > https://bugs.openjdk.java.net/browse/JDK-8146264
JSType and few other places have "instanceof Number" check -- not sure if fixing JSType.toStringImpl alone will do. JSType和其他一些地方都有“instanceof Number”检查 - 不确定是否单独修复JSType.toStringImpl。 In any case, I've a workaround - not very pretty one - but a workaround nevertheless. 无论如何,我有一个解决方法 - 不是很漂亮 - 但仍然是一个解决方法。 You can call java.lang.Object.toString method on those objects thereby avoiding Nashorn's JSType string conversion code. 您可以对这些对象调用java.lang.Object.toString方法,从而避免使用Nashorn的JSType字符串转换代码。
function fibonacci(n) {
var BigInteger = Java.type("java.math.BigInteger");
prev = new BigInteger("0");
if (n == 0) return prev;
next = new BigInteger("1");
if (n == 1) return next;
var i, fib = null;
for (i = 1; i < n; i++) {
fib = prev.add(next);
prev = next;
next = fib;
}
return fib;
}
function javaToString(obj) {
var javaToStringMethod = (new java.lang.Object()).toString;
var call = Function.prototype.call;
return call.call(javaToStringMethod, obj);
}
print(javaToString(fibonacci(77)))
print(javaToString(fibonacci(78)))
print(javaToString(fibonacci(79)))
var str, BigInteger = Java.type("java.math.BigInteger");
str = "9999999999999998";
print(str + ": " + javaToString(new BigInteger(str)));
str = "9999999999999999";
print(str + ": " + javaToString(new BigInteger(str)));
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.