## Simple recursive function not giving expected output

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Sorry if I did this wrong, i'm really new with C and haven't used stack overflow before. I'm trying to trace this simple recursive function by hand but am getting a different answer from the compiled code.

My thought process was

print 2 | n = 2-1 = 1 | 1 >= 0 | countdown(1)

print 1 | n = 1-1 = 0 | 0 >= 0| countdown(0)

print 0 | n = 0-1 = -1 | -1 is not >= 0|

print -1| END

```void countdown(int n)
{
printf ("n = %d\t", n);
n--;
if (n >= 0)
{
countdown(n);
}
printf ("n = %d\t", n);
}

int main ( )
{
countdown(2);
return 0;
}
```

I expected to get: n = 2 n = 1 n = 0 n = -1

but the compiled code gives me: n = 2 n = 1 n = 0 n = -1 n = 0 n = 1

I'm not quite sure where the additional 0 and 1 come from after -1

Your code does the following (omitting the if):

```countdown(n):
print(n)
countdown(n-1)
print(n-1)
```

For n=2:

```countdown(2):
print(2)
countdown(1)
print(1)

-> next recursion step
print(2)
print(1)
countdown(0)
print(0)
print(1)

-> next recursion step
print(2)
print(1)
print(0)
print(-1)
print(0)
print(1)
```

Python Tutorial: Recursive Functions, A recursion can end up in an infinite loop, if the base case is not met in the calls. Example: 4! = 4 * 3! 3! = 3 * 2! 2! = 2 * 1. Replacing the calculated values gives� Question: Tag: python,list,recursion I'm looking to reverse a list using recursion in python. I succeeded in doing so but the output isn't exactly what I wanted. trying to figure out where i'm going wrong.

Your code has no problem. Just remove 2nd `printf` code.

```void countdown(int n)
{
printf("n = %d\t", n);
n--;
if (n >= 0)
countdown(n);
}

int main()
{
countdown(2);
return 0;
}
```

The result is:

```n = 2 n = 1 n = 0
```

This is the callstack when I captured at `2nd printf`.

```StudyCpp.exe!countdown(int n) line 16   C++  // It is 2nd printf of countdown(0). Now, n is -1.
StudyCpp.exe!countdown(int n) line 14   C++  // It called countdown(0)
StudyCpp.exe!countdown(int n) line 14   C++  // It called countdown(1)
StudyCpp.exe!main() line 21 C++   // It called countdown(2)
```

If you progress debug one more, You can see the callstack as below:

```StudyCpp.exe!countdown(int n) line 16   C++  // It is 2nd printf of countdown(1) after executed countdown(0).
StudyCpp.exe!countdown(int n) line 14   C++  // It called countdown(1)
StudyCpp.exe!main() line 21 C++   // It called countdown(2)
```

And, If you progress debug one more, You can see the callstack as below:

```StudyCpp.exe!countdown(int n) line 16   C++  // It is 2nd printf of countdown(2) after executed countdown(1).
StudyCpp.exe!main() line 21 C++   // It called countdown(2)
```

And, Finally, Program will be exited.

Use recursion to create a range of numbers example, The function begins and n is not equal to 1 so it skips to last function call has an output, which is then used to the second last to give a value to because I would expect it to exit the function after returning the array once the� Each time the function is called, a new output list must be created. Then at line 31, you are computing the ancestors of the candidate by a recursive call but you are not doing anything with the list of ancestors that is returned by that call. If you don't use the result, this call is unnecessary, or perhaps there is something to change in the

Here you are calling countdown() recursively and is called thrice. Actually each recursive call will push countdown() into the stack.

```SF1(Bottom of stack) ---> SF2 ---> SF3 (Top of stack).
```

Now the frame in the top will be executed. During the return from the function, the particular stack frame will be popped out. Now Stack frame pointer points to SF2 and then SF1.

Considering your program, below is the flow. Push Operation: SF1 will be pushed onto the stack first

```n = 2 printed.
Again n updated to 1
```

SF2 pushed:

```n = 1 got printed.
Again n updated to 0
```

SF3 pushed:

```n = 0 got printed.
Again n updated to -1.
But n <= 0, So if check fails.
```

Pop operation: Now SF3 got popped out from stack first.

```n = -1 printed
```

Then SF2 popped

```prints n = 0.
```

Finally SF1

```n = 1 printed
```

JavaScript Recursive Function By Example, The recurse() is a recursive function if it calls itself inside its body, like this: Output: 3 2 1. The countDown() seems to work as expected. However, as mentioned in the Function type tutorial, the name of Uncaught TypeError: countDown is not a function Given a number e.g., 324, calculate the sum of digits 3 + 2 + 4 = 9. In this tutorial, you will learn to write recursive functions in C programming with the help of an example. A function that calls itself is known as a recursive function. And, this technique is known as recursion.

7.11 — Recursion, Consequently, it's rather simple to write a (not very efficient) recursive function to liking the quiz's that give a main () function and show the expected output and� How this C++ recursion program works As we can see, the factorial () function is calling itself. However, during each call, we have decreased the value of n by 1. When n is less than 1, the factorial () function ultimately returns the output.

[PDF] 16. Recursion, To illustrate the nature of recursion, consider again the function, factorial, which takes as an argument a Based on the pattern in these examples, we can say: Though that is the expected value, Java will not prevent a negative integer to This return value is received by main, which prints it, giving us the output: 2. Thus � Recursion is a programming term that means calling a function from itself. Recursive functions can be used to solve tasks in elegant ways. When a function calls itself, that’s called a recursion step. The basis of recursion is function arguments that make the task so simple that the function does not make further calls.

ReturnType Function(Pass appropriate arguments ) { if a simple case, return the simple value // base case / stopping condition. else call function with simpler version of problem} For a recursive function to stop calling itself we require some type of stopping condition. If it is not the base case, then we simplify our computation using the

• From the second `printf` in the function.