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AWS Lambda function can't read my .bin files (Java)

I wrote a program in Java and deployed it to AWS Lambda by creating a JAR file containing all the necessary files. My request handler calls a method that tries to create 4 new FileInputStreams, with .bin files as inputs, in a try-catch. For example: InputStream stream = new FileInputStream("random.bin"). However, it's going to the catch statement after trying to create the first one due to an IOException. I have my class files in src/main/java/package. I had the .bin files in the directory that contains the src folder, and I also tried putting them all in one folder, but that didn't resolve the problem. Very confused how to resolve this problem. It's also not working when I call to read in a text file, so it's not just limited to binaries. I was thinking of putting the binaries in my S3 bucket and somehow reading them from there, but I haven't found anything much online detailing how to do that in Java.
Thanks!

Your jar puts files in different locations then what your expect. This question might offer so insight on how to correctly access files in the jar.

AWS Lambda only allow write access in the /tmp folder of your function's filesystem. Try changing your code to match this path and try again.

Java - Unable to find a file path which certainly exists

I am sure this is a relatively simple question, and I actually think it may be more of a problem with Windows than with Java.
I have a method for copying a file to a new directory, which takes two File objects, a File created with the path of the original, and a File created with the desired path of the copy. I am sure that the method works because I have used it to successfully copy a file onto my Desktop.
However, using my actual desired path creates an error:
java.io.FileNotFoundException: PATH (The system cannot find the path
specified)
Where the PATH is the path that I am attempting to use.
Here is my guess:
I am making this program for use on another machine. As such, the path that I am trying to use is:
C:\Users\XXXXXX\rest_of_path\filename.file
where XXXXXX is the primary user on the machine which I am writing the program for.
This directory exists on my system, but XXXXXX is not a user on my system. So I am guessing that Windows is causing a problem because of that.
I'm now changing my code to use a solution which depends on the machine, and is not hardcoded (System.getProperty).
However, I'd really like to know why this problem is occurring, from an academic standpoint, as a Windows and Java user.
Thanks in advance.
EDIT: accidentally used forward slashes when I meant double backslashes. To ensure that it was not a spelling error, I simply copied the directory using windows, and pasted it into my program (then doubled up on the backslashes).
EDIT: several users have suggested something which is far more clean than what I am trying to do in the first place. I'm leaving this question open because I'm curious why it is not working.
EDIT: I used the solution above and I'm completely happy with it. I still don't know why Windows will not allow me to access the original path, but I guess I really don't care at this point. Thanks, everybody!

In java, and generally most programming languages, you don't always have to provide the exact directory of your file. Although it would be nice to see the code you're using to get the file, I'll provide how it can be done.
I'm assuming you aren't using new File("file.txt") because that retrieves files from the folder your program is in, and doesn't require an entire address like C:\...\...\.... You certainly don't want to use an entire address because different operating systems use different paths, obviously.
The best you can do is put your files and requested folders somewhere relative to your program is (whether it's class files or a .jar file).
But with Windows you can be sure that with System.getProperty("...") you can retrieve directory URLs as relative paths for your files/folders.
Documentation on System.getProperty here: http://docs.oracle.com/javase/tutorial/essential/environment/sysprop.html
I may not have helped or answered your question at all. But hopefully you'll find a solution.

What's the correct use of new Java 7 Path class?

I currently use this approach to load files in my applications:
Path path = Paths.get(ClassLoader.class.getResource(fileName).toURI());
where fileName is structured as "/package1/package2/folder/file.lol".
Are there better, simpler, or more correct methods to get a Path object?
I need that works inside JARs too.

If you want to access a file which is not in a jar, then there is better solutions. But since it seems that you want to access that is part of a jar, I would say no!

How to find file extension if the file has been renamed?

How to find file extension if the file has been renamed? Is there any tool are available for this?
Example: I have a file "1.doc"; I hope everybody know this is a Word document I just renamed as "1.txt". But the file is a Word document originally; how can I get the original file extension?

You cannot get the previous file extension back. All you can do is look at the contents of the file, and try to guess what it is.
Most (many?) file formats begin with a "magic" value that is intended to identify the file format, but this method can never be 100% accurate.
As #AaronDigulla mentionned in a comment, there are libraries out there that help you in the task of "guessing", and the answer(s) to that question contain pointers to some such libraries.

Once you rename the file, the previous name is not preserved.
As far as automatically guessing the format of the file from its contents, take a look at the Unix file utility. The version that's installed on my Ubuntu box lists Microsoft Word among the "magic" signatures that it knows about.

What you want to do is find out what its MIME type is. You can use this tutorial http://www.roseindia.net/java/java-get-example/get-mime-type.shtml

before renaming it save it in a separate string called originalName...because when you change the name of the string, the memory allocated is changed and theres is no go back or undo

Consider using javax.activation.FileDataSource from the JavaBeans Activation Framework. In particular use the getContentType() method.

Finding temporary file storage in Java

So I'm writing a Java application that uses Simple to store data as xml file, but it is hellishly slow with big files when it stores on a network drive compared to on a local hard drive. So I'd like to store it locally before copying it over to the desired destination.
Is there some smart way to find a temporary local file storage in Java in a system independent way?
E.g. something that returns something such as c:/temp in windows, /tmp in linux, and likewise for other platforms (such as mac). I could use application path but the problem is that the Java application is run from the network drive as well.

Try:
String path = System.getProperty("java.io.tmpdir");
See: http://java.sun.com/javase/6/docs/api/java/lang/System.html#getProperties%28%29
And to add it here for completeness sake, as wic mentioned in his comment, there's also the methods createTempFile(String prefix, String suffix) and createTempFile(String prefix, String suffix, File directory) methods from Java's File class.

System.getProperty("java.io.tmpdir")
The System and Runtime classes are those whose javadocs you should check first when something related to the system is required.

In the spirit of 'let's solve the problem' instead of 'let's answer the specific question':
What type of input stream are you using when reading into Simple? Be sure to use BufferedInputStream (or BufferedReader) - otherwise you are reading one byte/character at a time from the stream, which will be painfully slow when reading a network resource.
Instead of copying the file to local disk, buffer the inputs and you will be good to go.

try System.getProperty("java.io.tmpdir");