Question

与here一起玩并走到尽头。问题是Collectors.toList被不兼容类型 -

拒绝

/**
 * General pair of items.
 *
 * @param <P> - Type of the first item in the pair.
 * @param <Q> - Type of the second item.
 */
static class Pair<P, Q> {
    final P p;
    final Q q;

    public Pair(P p, Q q) {
        this.p = p;
        this.q = q;
    }

    @Override
    public String toString() {
        return "{" + p + "," + q + "}";
    }
}

/**
 * Gets the `n`th item is present in the array - otherwise returns null.
 *
 * @param a   - The array
 * @param n   - Which one in the array we want.
 * @param <T> - The type of the array entries.
 * @return - The `n`th entry in the array or null if not present.
 */
private static <T> T n(T[] a, int n) {
    return n < a.length ? a[n] : null;
}

/**
 * Pairs up each element in the arrays.
 *
 * @param ps  - The `P` array.
 * @param qs  - The `Q` array.
 * @param <P> - The type of the elements in the `P` array.
 * @param <Q> - The type of the elements in the `Q` array.
 * @return A list of `Pair`s of each element.
 */
static <P, Q> List<Pair<P, Q>> pairUp(P[] ps, Q[] qs) {
    return IntStream.range(0, Math.max(ps.length, qs.length))
            .mapToObj(i -> new Pair(n(ps, i), n(qs, i)))
            .collect(Collectors.toList());
}

/**
 * Splits the two strings on a separator and returns a list of Pairs of thje corresponding items.
 *
 * @param a         - The first string.
 * @param b         - The second string.
 * @param separator - The separator.
 * @return - A List of Paired up entries from `a` and `b`.
 */
private static List<Pair<String, String>> fold(String a, String b, String separator) {
    return pairUp(a.split(separator, -1), b.split(separator, -1));
}

public void test() {
    System.out.println(fold("1;2;3;4", "Value1;Value2;Value whitespace", ";"));
}

我是否将推理规则推得太远了？如何解决这个问题？

Answer 1

将菱形运算符添加到Pair构造函数

new Pair<>(n(ps, i), n(qs, i))

泛型类型推断不起作用？

1 个答案: