StackOverflow上面給出的解釋是：
The reason for the HotSpot JVM's two survivor spaces is to reduce the need to deal with fragmentation. New objects are allocated in eden space. All well and good. When that's full, you need a GC, so kill stale objects and move live ones to a survivor space, where they can mature for a while before being promoted to the old generation. Still good so far. The next time we run out of eden space, though, we have a conundrum. The next GC comes along and clears out some space in both eden and our survivor space, but the spaces aren't contiguous. So is it better to
- Try to fit the survivors from eden into the holes in the survivor space that were cleared by the GC?
- Shift all the objects in the survivor space down to eliminate the fragmentation, and then move the survivors into it?
- Just say "screw it, we're moving everything around anyway," and copy all of the survivors from both spaces into a completely separate space--the second survivor space--thus leaving you with a clean eden and survivor space where you can repeat the sequence on the next GC?
Sun's answer to the question is obvious.
對(duì)于如何達(dá)到“無碎片”的目的，理解上可能有些困難，下面我把新生代回收機(jī)制詳細(xì)解釋一下：
注意，兩個(gè)survivor是交替使用的，在任意一個(gè)時(shí)刻，必定有一個(gè)survivor為空，一個(gè)survivor中存放著對(duì)象（連續(xù)存放，無碎片）。回收過程如下：
S1、GC，將eden中的live對(duì)象放入當(dāng)前不為空的survivor中，將eden中的非live對(duì)象回收。如果survivor滿了，下次回收?qǐng)?zhí)行S2；如果survivor未滿，下次回收仍然以S1的方式回收；
S2、GC，將eden和存放著對(duì)象的survivor中的live對(duì)象放入當(dāng)前為空的survivor中，將非live對(duì)象回收。
可以看到，上述的新生代回收機(jī)制保證了一個(gè)survivor為空，另一個(gè)非空survivor中無碎片。
在執(zhí)行一定次數(shù)的minor GC后，會(huì)通過Full GC將新生代的survivor中的對(duì)象移入老年代。
對(duì)于理解GC的整個(gè)機(jī)制，推薦一篇非常好的文章：http://www.cubrid.org/blog/dev-platform/understanding-java-garbage-collection/