Iterating objects in a java program

Question

Please note the logger info messages and the values of the Longs. .getId() returns a Long value. procDefIdIn is defined as :

private static Long procDefIdIn = null;

procDefIdIn is passed to this class with a value of 131. How does the last line of log info occur? I also tried converting them to Strings using toString(). Same result.

private static void getDefinitions(List<DefinitionProcess> processes)
{
    Long procDefId = null;
    for (Iterator<DefinitionProcess> iterator = processes.iterator(); iterator.hasNext(); ) {
        DefinitionProcess proc = iterator.next();
        if(procDefIdIn != null){
            procDefId = procDefIdIn;

            logger.info(String.format("proc  = #%d procDefIdIn = #%d", proc.getId(), procDefIdIn));

            if(procDefIdIn != proc.getId()){ // How does 131 && 131 get past this?

                logger.info(String.format("Removed proc = #%d", proc.getId()));

                iterator.remove();

                continue;
            }
        }
        else{
            procDefId = proc.getId();
        }

          ........
          //Code here does not execute
         .......

    }
}

INFO - proc = #129 procDefIdIn = #131

INFO - Removed proc = #129

INFO - proc = #130 procDefIdIn = #131

INFO - Removed proc = #130

INFO - proc = #131 procDefIdIn = #131

INFO - Removed proc = #131 //?!?!

Sebri Zouhaier · Accepted Answer · 2015-08-11 14:29:49Z

3

Please use

if(!procDefIdIn.equals(proc.getId())){

answered Aug 11, 2015 at 14:29

Sebri Zouhaier

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3 Comments

Joe Essey Over a year ago

Use .equals() for non-primitive Object comparisons in Java, as this answer illustrates.

Sebri Zouhaier Over a year ago

Long is not a primitive Data Type

user2568374 Over a year ago

@Sebri Thanks for that and your careful edit after the post :). Yes, I keep forgetting things like that as I am transferring from C++ to Java.

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