如何检查数组元素是否为null以避免Java中的NullPointerException
我有一个部分未填充的对象数组，当我遍历它们时，我试图在对其执行其他操作之前检查所选对象是否为如何检查数组元素是否为null以避免Java中的NullPointerException,java,arrays,exception-handling,nullpointerexception,Java,Arrays,Exception Handling,Nullpointerexception,我有一个部分未填充的对象数组，当我遍历它们时，我试图在对其执行其他操作之前检查所选对象是否为null。然而，即使是检查它是否为null的行为也似乎是通过NullPointerException实现的数组。长度也将包括所有null元素。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 (so
nullnullNullPointerExceptionnullnullObject[][] someArray = new Object[5][];
for (int i=0; i<=someArray.length-1; i++) {
if (someArray[i]!=null) { //do something
}
}
Object[][]someArray=新对象[5][]；
对于（int i=0；i，您所做的事情比您所说的要多。我根据您的示例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");
}
}
}
}
您可能正在尝试检查someArray[index]的长度吗？给定的代码对我有效。请注意，someArray[i]始终为空，因为您尚未初始化数组的第二维度。首先，这些代码无法编译
在i++之后删除多余的分号后，它对我来说编译并运行良好。它没有
请参阅下面。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>
C:\oreyes\samples\java\arrays>type ArrayNullTest.java
公共类ArrayNullTest{
公共静态void main（字符串[]args）{
Object[]someArray=新对象[5][]；
对于（int i=0；ijavac ArrayNullTest.java）
C:\oreyes\samples\java\arrays>java ArrayNullTest
0处的元素为null
1处的元素为空
位于2的元素为空
3处的元素为空
4处的元素为空
C:\oreyes\samples\java\arrays>
示例代码不会抛出NPE。（在i++后面也不应该有“；”）与代码是否正在编译作斗争我想说创建一个包含六个值的数组5添加两个值并打印它们，您将得到这两个值，其他值为空。问题是虽然大小为5，但数组中有两个对象。如何查找数组中存在多少对象
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");
}
如果要查找特定的空元素，应如上所述使用for循环。publicstaticvoidmain（字符串s[]）
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;
}
{
int firstArray[]={2,14,6,82,22}；
int secondArray[]={3,16,12,14,48,96}；
int number=getCommonMinimumNumber（第一个数组，第二个数组）；
System.out.println（“编号为”+编号）；
}
公共静态int getCommonMinimumNumber（int firstSeries[]，int secondSeries[]
{
整数结果=0；
if（firstSeries.length！=0&&secondSeries.length！=0）
{
系列（第一系列）；
系列（第二系列）；
一：for（int i=0；i编号[i+1]）
{
温度=编号[i+1]；
编号[i+1]=编号[i]；
编号[i]=温度；
固定=假；
}
}
}
/*for（int i=0；i
您可以在一行代码中完成（无需数组声明）：
您的代码没有给我NPR。您可能还想使用“如果您选中！someArray.equals（null）
，iIt将抛出NPE”。
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;
}
object[] someArray = new object[]
{
"aaaa",
3,
null
};
bool containsSomeNull = someArray.Any(x => x == null);