Converting String<->byte[] -- No Data Loss?

[Hal Vaughan]

If I have a byte[] and I convert it to string (String sData = new
String(byte[] bData), then convert it back (byte bData = sData.getBytes( )),
will all data be intact, or do Strings have problems with bytes that are
not printable characters? I've tested this and it seems to work fine, but
I want to make sure there isn't some condition or situation I'm not aware
of that could cause problems.

I'm doing this because it's easier to do some of my work with strings, so
I'm reading a Zip file, saving each entry as a String, modifying some, then
saving it as a new Zip. I've found with Zip files, you can't always be
sure of how much data you'll be able to read in, so I'm reading in data in
blocks upto 2k, then taking each block and converting it to a String, so I
can concatenate the amount read in onto the end of another string. (With
Strings, I can use sString.substri ng(), I can't find a way to concatenate
byte[] arrays without continually creating a new array and using
System.arrayCop y().)

Thanks for any info.

Hal

[Chris]

Re: Converting String&lt;-&gt;byte[] -- No Data Loss?

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Hal Vaughan wrote:
[color=blue]
> If I have a byte[] and I convert it to string (String sData = new
> String(byte[] bData), then convert it back (byte bData =
> sData.getBytes( )), will all data be intact, or do Strings have
> problems with bytes that are
> not printable characters? I've tested this and it seems to work
> fine, but I want to make sure there isn't some condition or
> situation I'm not aware of that could cause problems.
>
> I'm doing this because it's easier to do some of my work with
> strings, so I'm reading a Zip file, saving each entry as a String,
> modifying some, then
> saving it as a new Zip. I've found with Zip files, you can't always
> be sure of how much data you'll be able to read in, so I'm reading
> in data in blocks upto 2k, then taking each block and converting it
> to a String, so I
> can concatenate the amount read in onto the end of another string.
> (With Strings, I can use sString.substri ng(), I can't find a way to
> concatenate byte[] arrays without continually creating a new array
> and using System.arrayCop y().)
>
> Thanks for any info.
>
> Hal[/color]

Hello,
Definitely don't do this. It is possible to lose data. Instead, what
you want to do, probably, is use a ByteArrayOutput Stream. Basically,
what you do, is something like this:

<Given that 'is' is an input stream reading from the ZIP entry>

ByteArrayOutput Stream bout = new ByteArrayOutput Stream();
byte[] buffer = new byte[2048];
int i;
while ((i = is.read(buffer) ) != -1) {
bout.write(buff er, 0, i);
}

byte[] entireEntry = bout.toByteArra y();

- --
Chris
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[Hal Vaughan]

Re: Converting String&lt;-&gt;byte[] -- No Data Loss?

Chris wrote:
[color=blue]
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>
> Hal Vaughan wrote:
>[color=green]
>> If I have a byte[] and I convert it to string (String sData = new
>> String(byte[] bData), then convert it back (byte bData =
>> sData.getBytes( )), will all data be intact, or do Strings have
>> problems with bytes that are
>> not printable characters? I've tested this and it seems to work
>> fine, but I want to make sure there isn't some condition or
>> situation I'm not aware of that could cause problems.
>>
>> I'm doing this because it's easier to do some of my work with
>> strings, so I'm reading a Zip file, saving each entry as a String,
>> modifying some, then
>> saving it as a new Zip. I've found with Zip files, you can't always
>> be sure of how much data you'll be able to read in, so I'm reading
>> in data in blocks upto 2k, then taking each block and converting it
>> to a String, so I
>> can concatenate the amount read in onto the end of another string.
>> (With Strings, I can use sString.substri ng(), I can't find a way to
>> concatenate byte[] arrays without continually creating a new array
>> and using System.arrayCop y().)
>>
>> Thanks for any info.
>>
>> Hal[/color]
>
> Hello,
> Definitely don't do this. It is possible to lose data. Instead, what
> you want to do, probably, is use a ByteArrayOutput Stream. Basically,
> what you do, is something like this:
>
> <Given that 'is' is an input stream reading from the ZIP entry>
>
> ByteArrayOutput Stream bout = new ByteArrayOutput Stream();
> byte[] buffer = new byte[2048];
> int i;
> while ((i = is.read(buffer) ) != -1) {
> bout.write(buff er, 0, i);
> }
>
> byte[] entireEntry = bout.toByteArra y();[/color]

I've never used (or heard of) ByteArrayOutput Stream. I looked it up. Am I
correct to say that what you're doing is writing to bout and that you can
keep writing to bout and it basically just keeps filling up until you're
done, then you can move the entire amount into a new byte array?

Thanks!

Hal
[color=blue]
> - --
> Chris
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[Chris]

Re: Converting String&lt;-&gt;byte[] -- No Data Loss?

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Hal Vaughan wrote:
[color=blue]
> I've never used (or heard of) ByteArrayOutput Stream. I looked it
> up. Am I correct to say that what you're doing is writing to bout
> and that you can keep writing to bout and it basically just keeps
> filling up until you're done, then you can move the entire amount
> into a new byte array?
>
> Thanks!
>
> Hal[/color]

Bingo. It gets as big as you need. I'm pretty sure you can even call
toByteArray(), use the results, and then (or even at the same time)
add MORE stuff to it and call toByteArray() again, returning
everything. It's also very useful if you need to generate some kind
of binary format and turn it into a byte array: you can just wrap it
with a DataOutputStrea m and use the methods provided there (if you
need to do this but you know the output size, I think that using
java.nio.ByteBu ffer and wrap()ing an array is probably faster but, as
I said, it requires that you know the size beforehand).

- --
Chris
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[Hal Vaughan]

Re: Converting String&lt;-&gt;byte[] -- No Data Loss?

Chris wrote:
[color=blue]
> -----BEGIN PGP SIGNED MESSAGE-----
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>
> Hal Vaughan wrote:
>[color=green]
>> I've never used (or heard of) ByteArrayOutput Stream. I looked it
>> up. Am I correct to say that what you're doing is writing to bout
>> and that you can keep writing to bout and it basically just keeps
>> filling up until you're done, then you can move the entire amount
>> into a new byte array?
>>
>> Thanks!
>>
>> Hal[/color]
>
> Bingo. It gets as big as you need. I'm pretty sure you can even call
> toByteArray(), use the results, and then (or even at the same time)
> add MORE stuff to it and call toByteArray() again, returning
> everything. It's also very useful if you need to generate some kind
> of binary format and turn it into a byte array: you can just wrap it
> with a DataOutputStrea m and use the methods provided there (if you
> need to do this but you know the output size, I think that using
> java.nio.ByteBu ffer and wrap()ing an array is probably faster but, as
> I said, it requires that you know the size beforehand).[/color]

When I first started learning Java (I'm entirely self taught), I found IO
streams frustrating, but now that I see all the things you can do -- and
how easy it is to wrap one stream around another to do some complex stuff
fairly easily, they make a lot more sense. It just takes time to get used
to them

Thanks!

Hal
[color=blue]
> - --
> Chris
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