Yes i understanding this issue large byte array size overflow how
handle here.give me way please or perfect url
On Fri, Sep 2, 2011 at 12:24 PM, Miguel Morales <therevolti...@gmail.com> wrote:
> Look up examples on stream reading a file for java.
> The problem is you are allocating an extremely large byte array.
>
> On Thu, Sep 1, 2011 at 11:51 PM, NaveenShrivastva
> <kumarnaveen.si...@gmail.com> wrote:
>> here i am taking video file for read byte the error byte arry outofmemory
>>
>> On Fri, Sep 2, 2011 at 12:13 PM, Naveen <kumarnaveen.si...@gmail.com> wrote:
>>> Please see this example when i am taking small file then it works but
>>> when taking large video size with 1gb approx then outofmemory error
>>>
>>>
>>> import java.io.*;
>>>
>>> public class ReadFileByteArray {
>>>
>>> public static void main(String[] args) {
>>>
>>> //create file object
>>> File file = new File("C://FileIO//video.mp4");
>>>
>>> try
>>> {
>>> //create FileInputStream object
>>> FileInputStream fin = new FileInputStream(file);
>>>
>>> /*
>>> * Create byte array large enough to hold the content of the file.
>>> * Use File.length to determine size of the file in bytes.
>>> */
>>>
>>>
>>> byte fileContent[] = new byte[(int)file.length()];
>>>
>>> /*
>>> * To read content of the file in byte array, use
>>> * int read(byte[] byteArray) method of java FileInputStream class.
>>> *
>>> */
>>> fin.read(fileContent);
>>>
>>> //create string from byte array
>>> String strFileContent = new String(fileContent);
>>>
>>> System.out.println("File content : ");
>>> System.out.println(strFileContent);
>>>
>>> }
>>> catch(FileNotFoundException e)
>>> {
>>> System.out.println("File not found" + e);
>>> }
>>> catch(IOException ioe)
>>> {
>>> System.out.println("Exception while reading the file " + ioe);
>>> }
>>> }
>>> }
>>>
>>>
>>> /*
>>> Output would be
>>> File content :
>>> This file is for demonstration of how to read file in byte array
>>> using Java FileInputStream.
>>> */
>>>
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>
>
>
> --
> ~ Jeremiah:9:23-24
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>
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