![]() ![]() In the worst-case, this is still at least 32 or 64 times reduction in the amount of memory needed. This is not an option in Java, but it may be possible to squirrel away this bit in fields used for other things. Typically, for low-level realizations of this technique, that bit will be stored in the pointer itself leading to no additional memory and constant memory overall. So we need to allocate an extra bit of information as we iterate. We don't know whether to correct the left branch, process this node, and continue with the right branch, or to correct the right branch and go to our parent. When we get to a leaf, we process it, then go to our parent and then we have a conundrum. Using the above example, instead of pushing nodes when we loop, we just need to remember our immediate parent, and at each iteration, we set the link we traversed to the current parent and then the current parent to the node we are leaving. In this technique, you encode the stack into the structure you are traversing, and by reusing the links you are traversing, you can do this with no or significantly less additional memory. Heap is typically much more plentiful than stack.Īs something that is usually an extremely bad idea, but is necessary in cases where memory use is extremely constrained, you can use pointer reversal. For an in order traversal you would loop down the left branches pushing each node as you visited it until you hit a leaf, which you would process, then pop a node off the top of the stack, process it, then restart your loop with the right child (by just setting your loop variable to the right node.) This will use a constant amount of stack by moving everything that was on the stack to the heap in the Stack data structure. to not recurse) but to instead allocate it in the heap.įor example, for an unbalanced tree traversal, you could store the nodes that will need to be revisited in a Stack data structure. BOUNCE TAILS 2 JAVA CODEIn the latter situation, you need to alter your code to not allocate the information on the stack (i.e. You likely either a) have a bug in your code leading to an infinite recursion which is usually quite easy to diagnose and fix, or b) have code which can lead to very deep recursions for example recursively traversing an unbalanced binary tree. The correct answer is the one already given. You really want to find the entire melody, since that's what's common to all the stack overflows with the same root cause. ![]() If you are faced with a stack overflow, then, you want to ignore the top of the stack, since that's just focusing on the specific note that exceeded your vocal range. In real life, the loop can be quite long, leading to dozens of potential points where the stack overflow can manifest itself. The "recursion" in this analogy was rather quick, just eight bars before the loop repeated. In other words, the same underlying runaway recursion (musically represented by an ever-higher rendition of the melody) can manifest itself in five different ways. If the melody represented a program's stack usage, a stack overflow could possibly occur at any of those five locations in the program's execution. In the melody, the first three notes are each a new "record high" (i.e., the notes are higher than any other note sung so far), and new record highs appear in the three notes of the third measure, and a final record high in the second note of the fifth measure. Eventually, you will reach the top of your singing range, and precisely where that happens depends on where your vocal limit lines up against the melody. ![]() Suppose you're singing the song Frère Jacques, except that you sing each verse a few tones higher than the previous one. That's because stack overflows tend to happen at a random point in the recursion each stack overflow looks superficially different from every other one even if they are the same stack overflow. If you go hunting through your defect tracking database trying to see whether this is a known issue or not, a search for the top functions on the stack is unlikely to find anything interesting. Take a look at Raymond Chen's post When debugging a stack overflow, you want to focus on the repeating recursive part. ![]()
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