递归函数导致堆栈溢出

Question

I have the following task:

Write a program that asks for a number and a power. Write a recursive function that takes the number to the power. For example, if the number is 2 and the power is 4, the function will return 16.

I wrote a program and there are no errors when I compile it, but when I start the program and enter a value gives an error saying "Stack Overflow". I suppose my recursive function became infinite but I have no idea how to write it in other way.

This is my code:

#include <iostream>
using namespace std;
int powpow(int number);

int main(){
    cout<<"Enter number:";
    int x;
    cin>>x;
    cout<<"The result of ("<<x<<" * "<<x<<") * "<<x*x<<" is: "<<powpow(x);
    system("pause");
    return 0;
}
int powpow(int number){
    int result = number*number;
    return powpow(result);
}

Answer 1

You have no terminating condition for your recursion, so it runs forever.

It sounds like maybe you don't have a good grasp of recursion, so I'd like to start with something a little simpler, the Fibonacci sequence .

Any time we define a function in terms of recursion, we need to first define a base case(s). In the case of Fibonacci, we have 2 base cases:

F(0) = 0
F(1) = 1

That says, in english, "F of 0 is 0, F of 1 is 1". Or even more simply, if we pass 0 to function F, we will get 0 back. If we pass 1, we will get 1 back.

Once we have the base cases defined, then we need to look for a recurrence relation. In the case of Fibonacci, we have the following recurrence:

F(n) = F(n-1) + F(n-2)

So for n >= 2 , we can use the above recurrence. Why? Well, lets try it for n = 2.

F(2) = F(n-1) + F(n-2)
     = F(1) + F(0)
     = 1    + 0
     = 1

So now we know that the answer to F(2) is 1. And what's more, we can now compute the answer to F(3). Why? Well, what do we need to compute F(3)? We need F(2) and F(1). We now have both of those answers since F(1) is a base case, and we just solved F(2) above.

So, now let's try to write a piece of pseudo code to solve F.

function F(int n) {
  // handle base cases
  if (n equals 0)
    return 0
  if (n equals 1)
    return 1

  // recurrence
  return F(n-1) + F(n-2);
}

Note that in a recursive function, we always handle the base cases at the beginning of the function. We cannot define this recurrence if we don't have base cases in place, otherwise, we will have no terminating condition for our recurrence. So that's why you always put the base cases at the beginning of the function.

Now, given the above explanation, another good exercise would be to write a recursive function for the factorial function . So, follow these steps:

1. Define the base case (use wikipedia article for hints).
2. Define recurrence in terms of base case
3. Write pseudo code to solve the recurrence, and be sure to put base case(s) at beginning of function, and recurrence at end.

Once you grasp these steps, then moving on to the power recurrence should make much more sense to you.

Answer 2

Your function

• does not what it should do
• has no termination condition

Try to think about this: How can your function return x^y when it only takes one number as a parameter. Then, think about how you raise number to a power and the implementation should be obvious.

Answer 3

Recursive routines always need a "trivial" or "base" case. Think about what you wrote, pass in 1 for x, what will the stop the recursion?

powpow(1)
  result = 1*1
  call powpow(1)
    result = 1*1
    call powpow(1)
      result = 1*1
      call powpow(1)

adinfinitum (or until you exeed the stack)