Java Map.put() throwing null pointer

Question

THIS IS NOT A DUPLICATE. It looks like it is, on the surface, but the solution for the other problems does not apply.

When I run the following method under Java 1.7, I get a null pointer exception at

normalOutCurrVals.put(new Double(x), new Byte(i));

I have read the documentation and the answers here for closely related problems, but so far I and my team are stumped. Is there something obvious that we are missing?

We simply wish to put in the mappings from double values (that appear on our programs UI) over to their byte counterparts for storage in a simulated device's memory.

Here are the importants declarations:

public final Map<Double, Byte>   normalOutCurrVals =   mapNormalOutCurrVals();
public final Map<Double, Byte>   magnetOutCurrVals = mapMagnetOutCurrVals();
public final Map<Integer, Byte>  freqVals          = mapFreqVals();
public final Map<String, Byte>   onTimeVals        = mapOnTimeVals();
public final Map<String, Short>  offTimeVals       = mapOffTimeVals();
public final Map<String , Short> pulseWidthVals    = mapPulseWidthVals();

Any help would be greatly appreciated.

private Map<Double, Byte> mapNormalOutCurrVals(){

    Map<Double, Byte> map = new HashMap<Double, Byte>();
    double x = 0.000;
    byte i = 0;

    //first set of normal output currents
    while (x <= 2.000){
        normalOutCurrVals.put(new Double(x), new Byte(i));
        x += 0.125;
        i++;
    }

    while (x <= 4.000){
        normalOutCurrVals.put(new Double(x), new Byte(i));
        x += 0.250;
        i++;
    }

    while (x <= 8.000){
        normalOutCurrVals.put(new Double(x), new Byte(i));
        x += 0.500;
        i++;
    }

    return map;
}

Why do you do `new Double(x)`? Autoboxing should take care of that. — tobias_k, Apr 13 '16 at 15:35
Why are you creating a `map` (and returning it empty), but calling put on `normalOutCurrVals`? — Mark Rotteveel, Apr 13 '16 at 15:36
So `normalOutCurrVals` uses itself during its initialization. You seem to have a typo. Use the new `map` instead. — Savior, Apr 13 '16 at 15:37
The solution does apply. It will always apply. There is only ever one reason for `NullPointerException`. Debug your code and find out what is `null`. (Hint: it's `normalOutCurrVals`.) — Savior, Apr 13 '16 at 15:39
@ tobias It doesn't auto-box. Eclipse insists that I explicitly convert to class Double — Jonah Havel, Apr 13 '16 at 15:40
@JonahHavel Then it looks you are using Java 1.4 source compatibility. — Mark Rotteveel, Apr 13 '16 at 15:41
have you tried i = (byte) (i + 1); instead of i++; I dont know if you can do it with i++ — Ricardo, Apr 13 '16 at 15:42
Two problems, you are returning an empty map, and in the private method you are accessing the public variable which hasn't been initialised. — jr593, Apr 13 '16 at 15:44
What a dork. I actually have several versions of this method, for different types of values. This one is the only one with the typo. Yet, I still didn't see it :( — Jonah Havel, Apr 13 '16 at 15:51

score 1 · Accepted Answer · answered Apr 13 '16 at 15:35

i think you want to fill your map!

private Map<Double, Byte> mapNormalOutCurrVals(){

Map map = new HashMap();
double x = 0.000;
byte i = 0;

//first set of normal output currents
while (x <= 2.000){
    map.put(new Double(x), new Byte(i));
    x += 0.125;
    i++;
}

while (x <= 4.000){
    map.put(new Double(x), new Byte(i));
    x += 0.250;
    i++;
}

while (x <= 8.000){
    map.put(new Double(x), new Byte(i));
    x += 0.500;
    i++;
}

return map;
}

Java Map.put() throwing null pointer

1 Answers1