Differentiate between two classes

Question

I have a class FileRecord and the class public class ZippedFileRecord : FileRecord, that get stored in SortedList mFiles which has all the FileRecords and ZippedFileRecords. The issue I'm running into is that I need a way to differentiate between the two, each ZippedFileRecord has a FileRecord and a method called GetFileRecord() where it unzips the file associated with ZippedFileRecord and makes a FileRecord for it (as seen in the chunk below..)

    public class ZippedFileRecord : FileRecord
    {
        FileInfo zipFileInfo; //FileName, FileCreationTime/Date
        public FileRecord myRecord = null;
        public FileRecord GetFileRecord()
        {
            if (myRecord == null)
            {
                //unzip and overwrite FileRecord
                UnZipFile();
                //return myRecord
                return myRecord;
                 }
                 else return myRecord;
            }
            ...
            ...
        }

now the portion of code that uses these FileRecords and ZippedFileRecords is something that a user can grab all the FileRecords in a certian date range, and get the info for those displayed on screen. Since ZippedFileRecords are representative of Zipped Files they must first be unzipped. here is the portion of code used to seek out the files to be displayed:

public GetFrames(DateTime desiredStartTime, DateTime  desiredEndTime)
{
    for(int fileIdx = mFiles.Values.Count-1; fileIdx >= 0; --fileIdx)
    {
        FileRecord rec = (FileRecord)mFiles.GetByIndex(fileIdx);
        if(rec.StartTime>= desiredStartTime && rec.EndTime<=desiredEndTime)
        {
            ...
        }
        else
        {
            ...
        }
    }
}

The Line that is causing me issues is FileRecord rec = (FileRecord)mFiles.GetByIndex(fileIdx); is there any way for me to check weather the mFiles[fileIdx] is a ZippedFileRecord or a FileRecord? because If I can do that I can unzip only as needed instead of unzipping and then rezipping potentially hundreds of Files every time they are to display on screen only to have one or two actually fit in the user's date range

Or the `is` operator http://msdn.microsoft.com/en-us/library/scekt9xw(v=vs.110).aspx — Mads, Jul 01 '13 at 19:46
FileRecord.GetFileRecord should be a virtual method overridden by ZippedFileRecord. That should elimate the issue. — P.Brian.Mackey, Jul 01 '13 at 19:47
@newStackExchangeInstance that doesn't help at all, I don't want to caste one as the other, I want to check which type each is... — Medic3000, Jul 01 '13 at 19:52
`as` isn't a cast, really. It's a special CLR thing. It's better than checking for `is` and then casting. — It'sNotALie., Jul 01 '13 at 19:54
`is` internally cast the reference to see if its null or not. So its always advised not to do `is` and then casting. — Rohit Vats, Jul 01 '13 at 20:01

Matthew Watson · Answer 1 · 2013-07-01T20:03:49.517

You can do:

var zippedFileRecord = mFiles.GetByIndex(fileIdx) as ZippedFileRecord;

if (zippedFileRecord != null)
    ... do something with it

It would be cleaner to do:

foreach (var zippedFileRecord in mFiles.OfType<ZippedFileRecord>())
    ... do something with zippedFileRecord

Although it seems that that you are doing something special by going backwards through the list, so this wouldn't work if that's the case.

But filtering by type like that is often a code smell. It may be that you should add a virtual GetFileRecord() method to class FileRecord implement it appropriately in that class. Then override it in class ZippedFileRecord to do the appropriate thing for that class.

Then you just call the GetFileRecord() method in the loop without needing to worry about the underlying type.

From the look of it, your implementation of GetFileRecord() in class FileRecord would simply be:

public virtual FileRecord GetFileRecord()
{
    return this;
}

And the implementation in class ZippedFileRecord would be the same as the one you already wrote, but with override added to the declaration:

public override FileRecord GetFileRecord()
{
    ...

Then you could use it like this:

for(int fileIdx = mFiles.Values.Count-1; fileIdx >= 0; --fileIdx)
{
    FileRecord rec = ((FileRecord)mFiles.GetByIndex(fileIdx)).GetFileRecord();
    ...

One last point: Why aren't you using the generic SortedList<TKey, TValue> so that you can use strong types to avoid casting?

This is the best answer. Small detail: if he's going backwards through the list, use `Reverse`. — It'sNotALie., Jul 01 '13 at 19:54

score 1 · Answer 2 · answered Jul 01 '13 at 19:46

1

You can test it like this:

if(mFiles.GetByIndex(fileIdx) is ZippedFileRecord)
{
   ...
}
else if(mFiles.GetByIndex(fileIdx) is FileRecord)
{
   ...
}

Note, that you need to check it in this order, since a ZippedFileRecord is also a FileRecord, but not vice versa.

answered Jul 01 '13 at 19:46

Romano Zumbé

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great! that was my big concern was with inheritance how I can check type – Medic3000 Jul 01 '13 at 19:50
@DarthSHeldon - You shouldn't need to. Re-check your design. Code is incomplete, but looks like you are method hiding. – P.Brian.Mackey Jul 01 '13 at 19:53

Rohit Vats · Answer 3 · 2013-07-01T19:55:59.140

1

Use the is operator -

if(rec is ZippedFileRecord)
{
  //Unzip logic here
}
else
{

}

OR

Make the method GetFileRecord() virtual in File record class and override it in ZippedFileRecord

public class FileRecord
{
    public virtual FileRecord GetFileRecord()
    {
       return this;
    } 
} 

public class ZippedFileRecord : FileRecord
{
    public override FileRecord GetFileRecord()
    {
       // Unzip and retunr FileRecord instance.
    } 
}

This way you don't have to worry for the type. This should work then -

FileRecord rec = mFiles.GetByIndex(fileIdx).GetFileRecord();

edited Jul 01 '13 at 19:55

answered Jul 01 '13 at 19:46

Rohit Vats

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Looks like we both came to the same conclusion. :) – Matthew Watson Jul 01 '13 at 20:02

score 0 · Answer 4 · answered Jul 01 '13 at 19:46

0

You could use is operator like this

if (!(rec is ZippedFileRecord)) 
{
   // do your stuff in case rec is not ZippedFileRecord
}

answered Jul 01 '13 at 19:46

Claudio Redi

67,454
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score 0 · Answer 5 · edited May 23 '17 at 11:56

0

You can use "is" to check if an object is of a specific type:

if (rec is ZippedFileRecord) {
    ...
}
else {
    ...
}

edit: See this question for more options.

edited May 23 '17 at 11:56

Community

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1

answered Jul 01 '13 at 19:46

Matthias

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score 0 · Answer 6 · answered Jul 01 '13 at 21:52

0

Personally, I'm wondering why ZippedFileRecord::GetFileRecord has to unzip the file. Since you only want the metadata, I'd refactor the method so that the metadata is set up without unzipping the file. Then, if you actually need the file data itself, then unzip the file.

answered Jul 01 '13 at 21:52

Eric Brown

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I don't need just the metadata, I also need all recorded data on the files – Medic3000 Jul 01 '13 at 22:04
But do you need all the data *at this point*? Laziness is a wonderful property. Compute the data you need *when you need it*. – Eric Brown Jul 01 '13 at 22:17

Differentiate between two classes

6 Answers6