如何检查数组元素是否为null以避免Java中的NullPointerException

Question

我有一个部分填充的对象数组，当我遍历它们时，我尝试检查所选对象是否为 null ，然后再对其进行其他操作。但是，即使是 null 检查它是否通过 NullPointerException 。 array.length 也将包含所有 null 元素。你如何检查数组中的 null 元素？例如，在下面的代码中将为我抛出一个NPE。

Object[][] someArray = new Object[5][];
for (int i=0; i<=someArray.length-1; i++) {
    if (someArray[i]!=null) { //do something
    } 
}

Answer 1

你比你说的更多。我从你的例子中运行了以下扩展测试：

public class test {

    public static void main(String[] args) {
        Object[][] someArray = new Object[5][];
        someArray[0] = new Object[10];
        someArray[1] = null;
        someArray[2] = new Object[1];
        someArray[3] = null;
        someArray[4] = new Object[5];

        for (int i=0; i<=someArray.length-1; i++) {
            if (someArray[i] != null) {
                System.out.println("not null");
            } else {
                System.out.println("null");
            }
        }
    }
}

并获得预期的输出：

$ /cygdrive/c/Program\ Files/Java/jdk1.6.0_03/bin/java -cp . test
not null
null
not null
null
not null

您是否可能尝试检查someArray [index]的长度？

Answer 2

没有。

见下文。您发布的程序按预期运行。

C:\oreyes\samples\java\arrays>type ArrayNullTest.java
public class ArrayNullTest {
    public static void main( String [] args ) {
        Object[][] someArray = new Object[5][];
            for (int i=0; i<=someArray.length-1; i++) {
                 if (someArray[i]!=null ) {
                     System.out.println("It wasn't null");
                 } else {
                     System.out.printf("Element at %d was null \n", i );
                 }
             }
     }
}


C:\oreyes\samples\java\arrays>javac ArrayNullTest.java

C:\oreyes\samples\java\arrays>java ArrayNullTest
Element at 0 was null
Element at 1 was null
Element at 2 was null
Element at 3 was null
Element at 4 was null

C:\oreyes\samples\java\arrays>

Answer 3

String labels[] = { "MH", null, "AP", "KL", "CH", "MP", "GJ", "OR" }; 

if(Arrays.toString(labels).indexOf("null") > -1)  {
    System.out.println("Array Element Must not be null");
                     (or)
    throw new Exception("Array Element Must not be null");
}        
------------------------------------------------------------------------------------------         

For two Dimensional array

String labels2[][] = {{ "MH", null, "AP", "KL", "CH", "MP", "GJ", "OR" },{ "MH", "FG", "AP", "KL", "CH", "MP", "GJ", "OR" };    

if(Arrays.deepToString(labels2).indexOf("null") > -1)  {
    System.out.println("Array Element Must not be null");
                 (or)
    throw new Exception("Array Element Must not be null");
}    
------------------------------------------------------------------------------------------

same for Object Array    

String ObjectArray[][] = {{ "MH", null, "AP", "KL", "CH", "MP", "GJ", "OR" },{ "MH", "FG", "AP", "KL", "CH", "MP", "GJ", "OR" };    

if(Arrays.deepToString(ObjectArray).indexOf("null") > -1)  {
    System.out.println("Array Element Must not be null");
              (or)
    throw new Exception("Array Element Must not be null");
  }

如果你想找到一个特定的null元素，你应该使用for循环，如上所述。

Answer 4

给定的代码适合我。请注意，someArray [i]始终为null，因为您尚未初始化数组的第二个维度。

Answer 5

好吧，首先，代码无法编译。

在i ++之后删除了额外的分号后，它编译并运行正常。

Answer 6

示例代码不会抛出NPE。 （也不应该是i ++背后的';'）。

Answer 7

争论代码是否正在编译我会说创建一个包含sixe 5的数组并添加2个值并打印它们，您将得到两个值，其他值为null。问题是虽然大小为5但阵列中有2个对象。如何查找数组中有多少个对象

Answer 8

public static void main(String s[])
{
    int firstArray[] = {2, 14, 6, 82, 22};
    int secondArray[] = {3, 16, 12, 14, 48, 96};
    int number = getCommonMinimumNumber(firstArray, secondArray);
    System.out.println("The number is " + number);

}
public static int getCommonMinimumNumber(int firstSeries[], int secondSeries[])
{
    Integer result =0;
    if ( firstSeries.length !=0 && secondSeries.length !=0 )
    {
        series(firstSeries);
        series(secondSeries);
        one : for (int i = 0 ; i < firstSeries.length; i++)
        {
            for (int j = 0; j < secondSeries.length; j++)
                if ( firstSeries[i] ==secondSeries[j])
                {
                    result =firstSeries[i];
                    break one;
                }
                else
                    result = -999;
        }
    }
    else if ( firstSeries == Null || secondSeries == null)
        result =-999;

    else
        result = -999;

    return result;
}

public static int[] series(int number[])
{

    int temp;
    boolean fixed = false;
    while(fixed == false)
    {
        fixed = true;
        for ( int i =0 ; i < number.length-1; i++)
        {
            if ( number[i] > number[i+1])
            {
                temp = number[i+1];
                number[i+1] = number[i];
                number[i] = temp;
                fixed = false;
            }
        }
    }
    /*for ( int i =0 ;i< number.length;i++)
    System.out.print(number[i]+",");*/
    return number;

}

Answer 9

你可以在一行代码上完成它（没有数组声明）：

object[] someArray = new object[] 
{
    "aaaa",
    3,
    null
};
bool containsSomeNull = someArray.Any(x => x == null);