Question: I have this line in my program :

```
InputStream Resource_InputStream=this.getClass().getClassLoader().getResourceAsStream("Resource_Name");

```

But how can I get FileInputStream from it [Resource\_InputStream] ?

Answer: 
Use [`ClassLoader#getResource()`](http://docs.oracle.com/javase/8/docs/api/java/lang/ClassLoader.html#getResource-java.lang.String-) instead if its URI represents a valid local disk file system path.

```
URL resource = classLoader.getResource("resource.ext");
File file = new File(resource.toURI());
FileInputStream input = new FileInputStream(file);
// ...

```

If it doesn't (e.g. JAR), then your best bet is to copy it into a temporary file.

```
Path temp = Files.createTempFile("resource-", ".ext");
Files.copy(classLoader.getResourceAsStream("resource.ext"), temp, StandardCopyOption.REPLACE_EXISTING);
FileInputStream input = new FileInputStream(temp.toFile());
// ...

```

That said, I really don't see any benefit of doing so, or it must be required by a poor helper class/method which requires `FileInputStream` instead of `InputStream`. If you can, just fix the API to ask for an `InputStream` instead. If it's a 3rd party one, by all means report it as a bug. I'd in this specific case also put question marks around the remainder of that API.
Long story short:
**Don't use FileInputStream** as a parameter or variable type. Use the abstract base class, in this case InputStream instead.