i want to read 1 kb data each time from byte array of 678560 size.

Question

1.67/5 (2 votes)

See more:

i want to read 1 KB data each time from byte array of 678560 size.

if in end 1 KB is not available then only read remaining bytes.

please help me.

Posted 18-Nov-13 20:06pm

bunty swapnil

Updated 18-Nov-13 21:36pm

Marco Bertschi

v3

Add a Solution

Comments

♥…ЯҠ…♥ 19-Nov-13 2:13am

what is the size = 678560b ?? 678560kb ?? 678560mb;

Sergey Alexandrovich Kryukov 19-Nov-13 2:46am

Help with what? Any particular problem? http://whathaveyoutried.com so far? You cannot expect that someone do all your work for you.
And what does it mean, "each time"? Each time of what?
—SA

2 solutions

Solution 2

Fredrik's solution is correct, but with just a few changes it can be generalized:

C#

public static class ChunkingExtensionMethod
{
  public static IEnumerable<IEnumerable<T>> Chunkify<T>(this IEnumerable<T> source, int chunkSize)
  {
    if (chunkSize <= 0)
      throw new ArgumentOutOfRangeException("chunkSize", "chunkSize must be greater than zero");
    while (source.Any())
    {
      yield return source.Take(chunkSize);
      source = source.Skip(chunkSize);
    }
  }
}

It will work with any source that implements IEnumerable<T> and doesn't make an array that it might not need.
Each of the elements of the returned enumeration can be converted to another collection type using .ToArray(), .ToList(), etc.

Use this like:

C#

var source = new byte[678560];
foreach (var chunk in source.Chunkify(1024))
{
  Console.WriteLine("Read {0} bytes IEnumerable", chunk.Count());
  Console.WriteLine("Read {0} bytes Array", chunk.ToArray().Length);
  Console.WriteLine("Read {0} bytes List", chunk.ToList().Count);
}

Posted 19-Nov-13 8:01am

Matt T Heffron

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Fredrik Bornander · Accepted Answer · 2013-11-18T21:08:00

Iterate over the source array of bytes, going by steps of 1024 bytes.
For each iteration, copy the data from the current point up until current+1024 into a target array, and while doing this make sure you don't copy past the end of the source buffer.

Using yield return and LINQ this can be achieved in using very little code:

C#

using System;
using System.Collections.Generic;
using System.Linq;

namespace Read1k {
    class Program {

        static IEnumerable<byte[]> SomeReadMethod(byte[] source, int chunkSize) {
            for (var i = 0; i < source.Length; i += chunkSize) {
                yield return source.Skip(i).Take(chunkSize).ToArray();
            }
        }

        static void Main() {
            var source = new byte[678560];
            foreach (var chunk in SomeReadMethod(source, 1024)) {
                Console.WriteLine("Read {0} bytes", chunk.Length);
            }
        }
    }
}

For each of the iterations done in Main, the chunk variable will hold a byte array with either 1024 bytes or whatever is left on the source array.

The LINQ methods Skip and Take will (obviously) skip ahead i bytes and then take the next chunkSize bytes.

Hope this helps,
Fredrik