Question

我有Guava Range上限和下限：

Integer upper = 5;
Integer lower = 3;
Range<Integer> range = Range.closed(lower,upper);

我希望在给定范围内的值类似于Math.min和Math.max：

Integer above = 6;
Integer within = 4;
Integer below = 2;

assert range.apply(above) == upper;
assert range.apply(below) == lower;
assert range.apply(within) == within;

Range.apply实现了Predicate.apply，所以上面的代码没有编译，但我正在寻找一些实用程序中的等效方法，否则会给出给定的值，如果它在边界内或返回最近的绑定端点，如果没有。

Answer 1

我编写了自己的实用程序方法：

/**
 * Ensures the value given is with the range or returns the nearest boundary endpoint.
 * 
 * WARNING: This only works with closed ranges.  
 * Open ranges will return an endpoint that will return false from Range.contains().
 * @param range
 * @param value
 * @return
 */
@Nonnull
public static <C extends Comparable<C>> C apply(@Nonnull Range<C> range, @Nonnull C value) {
    C result;
    if (value == null || range.contains(value)) {
        result = value;
    } else if (range.hasLowerBound() && range.lowerEndpoint().compareTo(value) >= 0) {
        result = range.lowerEndpoint();
    } else if (range.hasUpperBound() && range.upperEndpoint().compareTo(value) <= 0) {
        result = range.upperEndpoint();
    } else {
        throw new IllegalStateException("Range should contain or violate one of the bounds " + range + " for "
                + value);
    }
    return result;
}

我希望Guava已经提供了这种方法或其他替代方法。

应用番石榴范围最小/最大

1 个答案: