关于CharMatcher.WHITESPACE实现 [英] About the CharMatcher.WHITESPACE implementation

问题描述

当我查找 CharMatcher 的实现时，注意一个字段 WHITESPACE_MULTIPLIER = 1682554634 ，然后我设置了这个值到 1582554634 ，运行测试用例 CharMatcherTest＃testWhitespaceBreakingWhitespaceSubset ，当然失败了。

When i looked up the implementation of CharMatcher and notice a field WHITESPACE_MULTIPLIER=1682554634 , then i set this value to 1582554634 , running the testcase CharMatcherTest#testWhitespaceBreakingWhitespaceSubset, of course it failed.

之后我将testWhitespaceBreakingWhitespaceSubset更改为仅调用 WHITESPACE.apply（（char）c）而不断言，打印方法中的索引 WHITESPACE.matches

After that I changed testWhitespaceBreakingWhitespaceSubset to only invoke WHITESPACE.apply((char)c) without assert, print the index in the method of WHITESPACE.matches

int index=(WHITESPACE_MULTIPLIER * c) >>> WHITESPACE_SHIFT)

在更改了 WHITESPACE_MULTIPLIER 从 1682554634 到 1582554634

毫无疑问，1682554634设计得很好，我的问题是如何推断这个神奇的数字？

No doubt, 1682554634 is well designed , my question is how can I infer this "magic number"?`

在 Martin Grajcar的建议，我尝试编写魔术数字生成器，如下所示：

Upon Martin Grajcar's proposal, I try to write the "magic number generator" as follows and worked :

char[] charsReq = WHITESPACE_TABLE.toCharArray();
Arrays.sort(charsReq);
OUTER:
for (int WHITESPACE_MULTIPLIER_WANTTED = 1682553701; WHITESPACE_MULTIPLIER_WANTTED <= 1682554834; WHITESPACE_MULTIPLIER_WANTTED++) {
    int matchCnt = 0;
    for (int c = 0; c <= Character.MAX_VALUE; c++) {
        int position = Arrays.binarySearch(charsReq, (char) c);
        char index = WHITESPACE_TABLE.charAt((WHITESPACE_MULTIPLIER_WANTTED * c) >>> WHITESPACE_SHIFT);
        if (position >= 0 && index == c) {
                matchCnt++;
        } else if (position < 0 && index != c) {
                matchCnt++;
        } else {
            continue OUTER;
        }
    }
    // all valid
    if ((matchCnt - 1) == (int) (Character.MAX_VALUE)) {
        System.out.println(WHITESPACE_MULTIPLIER_WANTTED);
    }
}

如果更改了字符序列（swap \\\  WHITESPACE_TABLE中的算法没有解决方案（将循环结束条件更改为Integer.MAX_VALUE）。

if changed the sequence of characters(swap \u2001 \u2002 position) in WHITESPACE_TABLE the algorithms has no solution (changed the loop end condition to Integer.MAX_VALUE).

作为IntMath .gcd实现是指 http://en.wikipedia.org/wiki/Binary_GCD_algorithm

我的问题是：我在哪里可以找到 CharMatcher.WHITESPACE.match 实施的材料？

as the IntMath.gcd implementation is refer to http://en.wikipedia.org/wiki/Binary_GCD_algorithm
my question is : where can i find the material of CharMatcher.WHITESPACE.match implementation?

推荐答案

我不确定生成器是否仍然存在于某处，但可以轻松地重新创建。类结果包含 CharMatcher.WHITESPACE 的实施：

I'm not sure if the generator still exists somewhere, but it can be recreated easily. The class Result contains the data used in the implementation of CharMatcher.WHITESPACE:

static class Result {
    private int shift;
    private int multiplier;
    private String table;
}

// No duplicates allowed.
private final String allMatchingString = "\u2002\r\u0085\u200A\u2005\u2000"
        + "\u2029\u000B\u2008\u2003\u205F\u1680"
        + "\u0009\u0020\u2006\u2001\u202F\u00A0\u000C\u2009"
        + "\u2004\u2028\n\u2007\u3000";

public Result generate(String allMatchingString) {
    final char[] allMatching = allMatchingString.toCharArray();
    final char filler = allMatching[allMatching.length - 1];
    final int shift = Integer.numberOfLeadingZeros(allMatching.length);
    final char[] table = new char[1 << (32 - shift)];
    OUTER: for (int i=0; i>=0; ++i) {
        final int multiplier = 123456789 * i; // Jumping a bit makes the search faster.
        Arrays.fill(table, filler);
        for (final char c : allMatching) {
            final int index = (multiplier * c) >>> shift;
            if (table[index] != filler) continue OUTER; // Conflict found.
            table[index] = c;
        }
        return new Result(shift, multiplier, new String(table));
    }
    return null; // No solution exists.
}

它生成一个不同的乘数，但这没关系。

It generates a different multiplier, but this doesn't matter.

如果没有针对给定 allMatchingString 的解决方案，您可以减少转移并重试。

In case no solution for a given allMatchingString exists, you can decrement shift and try again.

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