Strings are wrappers around a char[]; String.substring will create a new String object with the *same* char[] from the original String. Your first snippet causes OutOfMemory because you're retaining the 100000 char Strings in someStringList even when you call substring. I don't have the Android implementation of String in front of me but I believe it's similar to the Sun implementation which merely passes the char[] to the new String.
Looking at the Javadocs for Android's String.substring, I must admit it could be easily misunderstood that a new String with new backing char[] is created. Your second snippet does not cause OutOfMemoryError because of the String concatenation. String concatenation creates a new String with a new backing char[]. The original char[100000] is garbage collected in that case. If you need the first char of each large String you could just create an ArrayList<Character> and just store String.charAt(0). If you need some range of the large String that's greater than one char, maybe you can create a StringBuilder/StringBuffer and append the substring to that, then store the toString() of that into your ArrayList. For example: String someBigString = new String(new char[100000]); someStringList.add(new StringBuilder(someBigString.substring(0, 1)).toString()); There's probably a better way to do this but I just got home from work and I'm not quite in a Java thinking mood. :-P Hope this helps. Ernest Woo Woo Games http://www.woogames.com On Oct 6, 2:50 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? -- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to android-developers@googlegroups.com To unsubscribe from this group, send email to android-developers+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/android-developers?hl=en
