CODEWITHSUNDEEP
javatestingcompilationnull
Dennis
Dennis

Reputation: 757

What's wrong with this? I'm getting symbol not found

This is the top four lines of the code method of a class:

This is IN the class 'CrudController'

public String create(Object entity, Class clazz, String req_id, HttpServletResponse response) throws NullPointerException {
    if (entity == null ) throw NullPointerException("ENTITY");
    else if(clazz == null) throw NullPointerException("CLAZZ");
            else if(response == null) throw NullPointerException("RESPONSE");

I'm getting these errors for these three lines:

[ERROR] /path/to/CrudController.java:[42,29] error: cannot find symbol
[ERROR]  class CrudController
[ERROR] /path/to/CrudController.java:[43,31] error: cannot find symbol
[ERROR]  class CrudController
[ERROR]/pathto/CrudController.java:[44,48] error: cannot find symbol
[ERROR]  class CrudController

The postion numbers (29, 31, and 48 are RIGHT at the beginning of '=='

Upvotes: 1

Views: 1241

Answers (2)

Vallabh Patade
Vallabh Patade

Reputation: 5110

change all throw NullPointerException(STRING); to throw new NullPointerException(STRING);

Here you are throwing the exception. Exception is instance of corresponding Exception type class so you need to throw an instance that why new keyword is needed.

public String create(Object entity, Class clazz, String req_id, HttpServletResponse     response) throws NullPointerException {
if (entity == null ) throw new NullPointerException("ENTITY");
else if(clazz == null) throw new NullPointerException("CLAZZ");
        else if(response == null) throw new NullPointerException("RESPONSE");

Upvotes: 5

cmd
cmd

Reputation: 11841

throw NullPointerException("CLAZZ"); is not a legal statement. You must use the new operator to create an instance of the NullPointerException

e.g.

throw new NullPointerException(STRING);

Upvotes: 0

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