On Thursday, October 7, 2010, DanH <danhi...@ieee.org> wrote:
> Yep, substring MAY (or may not) decide to avoid copying the characters
> and simply keep a reference to the char array in the source String. A
> smart substring implementation wouldn't do this if the result string
> were quite short or the source string quite long, but apparently the
> Android implementation isn't that smart.
It's a very reasonable behaviour of String class when we take into
account that String is immutable and very often Strings are
responsible for big chunks of memory allocation. Behaviour of
String.substring() makes therefore sense in some(most?) situations.
Any other behaviour could cause equally puzzling results in other
situations. Another important thing to take into account is that
String behaviour is imposed on JVM level (at least Sun's JVM but
Dalvik deriving from Apache Harmony will most likely implement similar
strategies).
Fault lies in details like these not being communicated well enough in
the documentation.
>
> The same thing may occur if you do "new String(someOtherString)".
>
> There's probably no guaranteed way to prevent this from happening,
> unless you somehow use a char array as an intermediate value.
>
> On Oct 6, 1:50 pm, Alex <a...@appfactory.at> wrote:
>> Hi everybody!
>> First of all, sorry for my bad english ;-)
>>
>> After several hours of searching for the cause of a OutOfMemoryError,
>> I found a wierd problem with Strings. To keep it simple, trimmed it
>> down to something like this:
>>
>> ArrayList<String> someStringList = new ArrayList<String>(1000);
>> for (int i=0; i<1000; i++) {
>> String someBigString = new String(new char[100000]);
>> String someSmallString = someBigString.substring(0,1);
>> someStringList.add(someSmallString);
>>
>> }
>>
>> OK, it's not very useful and even though it's a big string, I would
>> expect it to work properly, because only one char is stored in the
>> list. But in fact, heap is growing rapidly and OutOfMemoryError
>> exception will be thrown in 2,3 seconds. And I can't imagin why.
>>
>> AND NOW the interessting part, a little workaround:
>>
>> ArrayList<String> someStringList = new ArrayList<String>(1000);
>> for (int i=0; i<1000; i++) {
>> String someBigString = new String(new char[100000]);
>> String someSmallString = someBigString.substring(0,1);
>> someStringList.add(someSmallString + "BUGFIX");
>>
>> }
>>
>> WTF? Why is this now working?
>> The only difference is the concatenated string (someSmallString +
>> "BUGFIX")! And as originally expected, with this workaround nearly no
>> memory will be used ...
>>
>> Any ideas? I would appreciate it very much, if someone could give me
>> hint. Maybe I'm missing something basic here?
>
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