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up vote 0 down vote favorite

I am storing a double value inside the a HashMap as shown 

HashMap listMap = new HashMap();

double mvalue =0.0;

listMap.put("mvalue",mvalue );

Now when i tried to retrieve that value , as shown 

mvalue = Double.parseDouble((String) listMap.get("mvalue"));

i am getting an error as

java.lang.Double cannot be cast to java.lang.String

I am confused here ,

This is my actual HashMap and i am setting the values in it as shown 

HashMap listMap = new HashMap();
double mvalue =0.0 ;
List<Bag> bagList = null; 
listMap.put("bagItems",bagList);
listMap.put("mvalue", mvalue);

Could anybody please tell me , how the structure of the HashMap should be ?

share|improve this question
    
In addition to the answers to your question, you may also want to read the tutorial on generics. docs.oracle.com/javase/tutorial/java/generics/index.html –  Dunes Mar 6 '12 at 11:47

7 Answers 7

up vote 3 down vote

You have put a Double in the Map. Don't cast to String first. This will work:

HashMap<String, Double> listMap = new HashMap<String, Double>();
mvalue = listMap.get("mvalue");

Your primitive double is being Autoboxed to a Double Object. Use Generics to avoid the need to cast, which is the <String, Double> part.

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up vote 1 down vote

how about 

mvalue = listMap.get("mvalue");

?

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up vote 1 down vote

If you introduce generics you will immeditely discover the probblem

Map<String, Double> listMap = new HashMap<String, Double>();
Double mvalue = listMap.get("mvalue");

The reason is that your map does return a Double and not a String, hence the error message.

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up vote 0 down vote

First of all, make a HashMap with Class Types defined , like : HashMap<String,Double>

Secondly , this should work : 

mvalue = Double.parseDouble(Double.toString(listMap.get("mvalue")));

Although, best way is : 

mvalue = listMap.get("mvalue");
// provided defined HashMap is HashMap<String,Double>
share|improve this answer
    
if map stores double value then why to parse it? Here you are doing 2 unnecessary parsing. first double to string and then again string to double. –  Harry Joy Mar 6 '12 at 11:45
    
Sure, it works, but why go through the trouble of converting back and forth between String and Double when the value is already a Double? –  claesv Mar 6 '12 at 11:46
    
yeah i knw..And i modified my answer to suit its needs..but i wanted to tell @yyy i 777 , how to go about in this situation..!! –  Ramandeep Singh Mar 6 '12 at 11:47
    
The Map already holds a Double. If mvalue needs to be a primitive then unboxing will do it. Why on earth would you do the toString call, only to parse it as a Double ? –  planetjones Mar 6 '12 at 11:48
up vote 0 down vote

try this :

HashMap<String, Double> listMap = new HashMap<String, Double>();
Double mvalue = 0.0;
listMap.put("mvalue", mvalue);
mvalue = listMap.get("mvalue");
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up vote 0 down vote

Your code should looks like:

Map<String,Double> listMap = new HashMap<String,Double>();
double mvalue = 0.0;
listMap.put("mvalue", mvalue );
mvalue = Double.parseDouble( String.valueOf(listMap.get("mvalue")) );

Notice that if you need to retrieve a wrapper and not a primitive type, you can use on wrapper classes XXX.valueOf() instead of XXX.parseXXX() which returns a primitive.

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up vote 0 down vote

I guess your map will store many diferents types So I recomend <String, Object> generic

HashMap listMap = new HashMap<String, Object>();
double mvalue =0.0;
listMap.put("mvalue",mvalue );
.
.
.
String mValueString = Double.toString((Double) listMap.get("mvalue"));

This will get you the double object, cast it to Double, and the convert into string in new variable.

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