Question

class Program  
{
    static void Main(string[] args)
    {
        Test(0);
    }

    static void Test(int i)
    {
        if (i > 30000)
        {
            return;
        }
        Test(i + 1);
    }
}

Why getting recursive function and throwing StackOverflowException when calling this above sample.

(Because over the default recursive stack size ?)

but i wonder how to solve this problem .

Thanks.

Answer 1

You’re getting an exception because 30,000 stack frames is a rather large number 🙂

You solve it by using recursion in a more circumspect manner. The ideal problems to be solved recursively are those which reduce the “search space” quickly ^(a).

For example, binary tree traversal where your search space is halved each time you recur:

def find (searchkey, node):
    if node = NULL:
        return NULL
    if searchkey = node.key:
        return node
    if searchkey < node.key:
        return find (searchkey, node.left)
    return find (searchkey, node.right)

Adding two unsigned integers (and your own algorithm above) are not a good fit for recursion because you’ll blow out your stack allocation long before the results are calculated.:

def add (a, b):
    if a = 0:
        return b
    return add (a-1, b+1)

---

^(a) Search space can be defined as your entire set of possible answers. You want to reduce it as quickly as possible.

And , as an aside, the ideal problems for recursion have nothing to do with stack space in a theoretical/mathematical sense, they’re just any problem that can be expressed as:

• the same or similar problem with a “simpler” argument.
• a terminating condition with the “simplest” argument.

(“simple” , in this sense, means approaching the terminating condition).

Theoretical/mathemeatical approaches wouldn’t need to take stack space into account but we, as computer scientists, must. Reality sets limits 🙂

See also When would I not use recursion? and Situations where you would convert recursion to iteration.