C/C++: Compressing(gz) data without using temporary file

Question

I have some data in an array,

uchar *data = "Data to be compressed."; [Some data]

I want to compress(gz) this data and get output in a buffer.

uchar gzData[MAX_LEN];

One way to do this is:

 1. Creating a `ogzstream` object.
 2. Writing this data into a temporary file
 3. close the temporary file.
 3. Create a `ifstream` object.
 4. Read the temporary file into buffer using ifstream object .

But I do not want to use a temporary file for this as it is slowing down the whole process.

Is there any direct method of compressing this data without using any intermediate file ??

you do realize that what you are creating is a stream and not a file? There is filesystem interaction here. — niklasfi, Feb 17 '14 at 10:08

score 0 · Answer 1 · answered Feb 17 '14 at 10:09

0

You can use a filtering streambuf to do this on the fly. It's pretty easy to do, but Boost::Iostreams has one already done for you.

answered Feb 17 '14 at 10:09

James Kanze

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I do not know "Boost". :( Can u send me a link to some tutorial/example for the same? – techbull Feb 17 '14 at 10:19
How about std::stringstream? – Arsenii Fomin Feb 17 '14 at 11:00
You can start at www.boost.org. They've got a number of libraries; they tend to be overly generic (and thus overly complicated), but depending on what you're doing, some can be quite useful. – James Kanze Feb 17 '14 at 11:00
@FominArseniy `std::stringstream` doesn't support any compression. – James Kanze Feb 17 '14 at 11:06
@JamesKanze: Thanks for the suggestions sir. Actually, This is what I was actually trying to do http://stackoverflow.com/questions/7540259/deflate-and-inflate-zlib-h-in-c . – techbull Feb 18 '14 at 06:53

C/C++: Compressing(gz) data without using temporary file

1 Answers1