我如何检查空指针异常

Question

Here is the code

private static HashMap naturalNumbers = new HashMap();

static
{
    naturalNumbers.put("zero", new Integer( 0 ) );
    naturalNumbers.put("one", new Integer( 1 ) );
    naturalNumbers.put("two", new Integer( 2 ) );
    naturalNumbers.put("three", new Integer( 3 ) );
}

private static int findANumber( String partOfaNumber ) throws Exception
{
int multiplicand = 0;  
multiplicand += (Integer)naturalNumbers.get( partOfaNumber );

If the "get" returns null, how do I check for this?

I have tried:

if ( (Integer)naturalNumbers == null )
    {
        throw new Exception( "Number not found" );
    }
 return multiplicand;
}

but the IDE does not even accept it: cannot convert from HashMap to Integer.

Answer 1

Here's a slightly different version:

private static final HashMap<String, Integer> NUMS = new HashMap<String, Integer>();

static
{
    NUMS.put("zero", 0);
    NUMS.put("one", 1);
    NUMS.put("two", 2);
    NUMS.put("three", 3);
}

private static int findANumber(final String partOfaNumber) throws IllegalArgumentException
{
    int multiplicand = 0;  
    final Integer theNum = NUM.get(partOfaNumber);
    if (theNum != null) {
        multiplicand += theNum;
    } else {
        throw new IllegalArgumentException("Number not found (" + partOfNumber + ")");
    }

    return multiplicand;
}

Answer 2

This is simple.

Integer number = naturalNumbers.get( partOfaNumber )
if(number ==null) {
    throw new Exception("Number not found");
} else {
    multiplicand += number;
}
return multiplicand;

Answer 3

You can use shorthand as follows -

multiplicand += naturalNumbers.get(partOfaNumber ) == null ? 0 : (Integer) naturalNumbers.get(partOfaNumber );

Answer 4

private static int findANumber( String partOfaNumber ) throws Exception
{
  int multiplicand = 0;  
  if (naturalNumbers.containsKey( partOfaNumber ); {
    multiplicand += (Integer)naturalNumbers.get( partOfaNumber );
  } else {
    throw new Exception("Number not found");
  }
  return multiplicand;
}

Answer 5

Just check with the key if it is exist inside Hashmap. You dont have to worry about Null-Pointer Exception.

if(naturalNumbers.containsKey( partOfaNumber )){
  //Do your stuff
}else{
  //Do some stuff what you will when exception throw.
}

You can also search by Value by using

naturalNumbers.containsValue( value)

Answer 6

You will always have to add check for null values from collections for primitive types because of auto-boxing mechanism on java

A simple null pointer can be generated like below

Integer i = null;
i=i+10;

So always add default check for null and in your case you can add it like below

Integer multiplicand = 0;
Integer value = get(key);

if(value!= null)
{
//add value to multiplicand
}

Answer 7

First, it would be better to use a specific HashMap

private static HashMap<String, Integer> naturalNumbers = new HashMap<String, Integer>();

which will get rid of the nasty casting.

As for your problem, you try to cast the hash map to integer, which is not what you want. You want to check if a particular value exist. The better way is to extract the value and check for null:

int multiplicand = 0;
Integer part = naturalNumbers.get(partOfNumber)

if(part == null) {
   // Handle the null case
}

multiplicand += part;

Now, the question is, what are you going to do in the case the value doesn't exist. If you define the not existing value a zero, you can do do something like this

multiplicand += (naturalNumbers.contains(partOfNumber)?naturalNumbers.get(partOfNumber):0;

This makes it more concise. However, that depends on what you want to do if the value is not contained in the HashMap.