今天在生产环境联调的时候,发现一个很奇怪的问题,明明测试数据正确,结果却是结果不通过,经过debug查询到原来是Arrays.binarySearch用法错误,记录一下,避免后续再次犯错
具体测试如下:
想通过判断J是否存在数组中,结果发现出现如下错误
public static void main(String[] args) {
String[] test ={"X","J","7","5","4","11","W8","W7"};
System.out.println("没排序结果 = [" + Arrays.binarySearch(test,"J") + "]");
}
运行结果:
没有排序 = [-7]
进行排序后判断:
public static void main(String[] args) {
String[] test ={"X","J","7","5","4","11","W8","W7"};
System.out.println("没排序结果 = [" + Arrays.binarySearch(test,"J") + "]");
Arrays.sort(test);
System.out.println("排序后结果 = [" + Arrays.binarySearch(test,"J") + "]");
}运行结果:
没排序结果 = [-7]
排序后结果 = [4]
在网上查询下具体原因及查询官方解释如下:
/**
* Searches the specified array for the specified object using the binary
* search algorithm. The array must be sorted into ascending order
* according to the
* {@linkplain Comparable natural ordering}
* of its elements (as by the
* {@link #sort(Object[])} method) prior to making this call.
* If it is not sorted, the results are undefined.
* (If the array contains elements that are not mutually comparable (for
* example, strings and integers), it <i>cannot</i> be sorted according
* to the natural ordering of its elements, hence results are undefined.)
* If the array contains multiple
* elements equal to the specified object, there is no guarantee which
* one will be found.
*
* @param a the array to be searched
* @param key the value to be searched for
* @return index of the search key, if it is contained in the array;
* otherwise, <tt>(-(<i>insertion point</i>) - 1)</tt>. The
* <i>insertion point</i> is defined as the point at which the
* key would be inserted into the array: the index of the first
* element greater than the key, or <tt>a.length</tt> if all
* elements in the array are less than the specified key. Note
* that this guarantees that the return value will be >= 0 if
* and only if the key is found.
* @throws ClassCastException if the search key is not comparable to the
* elements of the array.
*/
public static int binarySearch(Object[] a, Object key) {
return binarySearch0(a, 0, a.length, key);
}
使用二分搜索法来搜索指定数组，以获得指定对象。在进行此调用之前，必须根据元素的自然顺序对数组进行升序排序（通过 sort(Object[]) 方法）。如果没有对数组进行排序，则结果是不确定的。（如果数组包含不可相互比较的元素（例如，字符串和整数），则无法 根据其元素的自然顺序对数组进行排序，因此结果是不确定的。）如果数组包含多个等于指定对象的元素，则无法保证找到的是哪一个,故所以会出现此问题
如果要判断数组中是否存在,可以使用 ArrayUtils.contains这个方法来判断是否存在,
/**
* 校验服务类型是否符合枚举值
* @param possibility 可能性
* @param field 字段
* @param info 校验不通过的错误消息
*
*/
public void validatePossibility(String[] possibility,String field,ErrorInfo info){
boolean flag = ArrayUtils.contains(possibility,field);
if (!flag) {
throw getException(info) ;
}
}