What Causes StackOverflowError?

Direct Answer

StackOverflowError means a thread's computation requires a larger JVM stack than the runtime permits. The usual cause is infinite or unexpectedly deep recursion, though a very deep call chain or unusually large frames can contribute. Fix the call structure or bound the input before treating stack-size tuning as the solution.

Deep Dive & Architecture

Recursive progress and the base case

A recursive method needs a base case and a step that moves every path toward it. visit(node.left) and visit(node.right) terminate for a true tree because each call moves to a smaller subtree and null is the base case. If the input is actually a graph with a back edge, the same code can revisit a node forever unless it keeps a visited set.

Repeated stack frames in the trace

Direct recursion repeats one method name. Mutual recursion repeats a small cycle such as parseExpression → parseTerm → parseExpression. The first useful frame in application code often identifies the input and condition that stopped progress. Do not hide the evidence by broadly catching Error.

Structural fixes for excessive recursion

Correct a missing base case, enforce a nesting limit, detect cycles, or replace recursion with an explicit work stack. The iterative form still uses O(depth) auxiliary storage, but depth becomes data that code can inspect and limit. Increasing -Xss may support a known bounded algorithm; it only delays failure for unbounded calls and consumes more memory per platform thread.

Two methods can overflow at different depths because their frames, runtime, platform, and compilation differ. Published universal depth or default-stack numbers are unreliable. Virtual threads change stack representation and mounting but do not make infinite recursion safe.

Code Example

class StackOverflowCause {
    static int sumTo(int n) {
        if (n < 0) throw new IllegalArgumentException("n must be non-negative");
        if (n == 0) return 0;
        return n + sumTo(n - 1);
    }

    public static void main(String[] args) {
        System.out.println(sumTo(10));
    }
}

Changing n - 1 to n + 1 removes progress and demonstrates the bug conceptually; do not run the broken form in a shared service.