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Editorial Team
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Asked: 2026-05-28T04:53:39+00:00

I have a Java app that fetches a relatively small .zip file using a

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I have a Java app that fetches a relatively small .zip file using a URL, saves it in a temp directory, unzips it onto the local machine and makes some changes to one of the files. This all works great.

However, I am accessing the .zip file via a BufferedInputStream in the following way:

Url url = "http://somedomain.com/file.zip";
InputStream is = new BufferedInputStream(url.openStream(), 1024);

My concern is that this app will actually be used to transfer very large zip files and I was wondering if a BufferedInputStream is actually the best way to do this, or whether I would just end up throwing some type of OutOfMemoryException?

So my question is, will a BufferedInputStream be suitable for this job, or should I be going about it in a completely different way?

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1 Answer

  1. Editorial Team

    BufferedInputStream doesn’t load all the file into memory, it only uses an internal buffer, in your case of size 1024 bytes = 1kb. It never gets larger than that. You could actually increase the value if you aren’t going to have many streams at once.

    Edit: what you are thinking about maybe is a ByteArrayOutputStream, where data is saved in memory.

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