CODEWITHSUNDEEP
cfunctionrecursion
Hannah_McDermott
Hannah_McDermott

Reputation: 17

Recursive function to print numbers counting down in c

I am tyring to write a program that reads an integer n from the keyboard, and prints n, n-1, ... n, using a recursive function. The code below works, but I don't completely understand why or the steps involving the recursive function. Could someone please explain?

void countingdown(int n){
    if (n == 1){
        printf("%d\n", n);
        return;
    }
    else {
        printf("%d\n", n);
        countingdown(n - 1);
        return;
    }
}

int main(){
    int n;
    printf("Enter an integer: ");
    scanf("%d", &n);
    countingdown(n);
    printf("\n");
    return 0;
}

Upvotes: 0

Views: 2330

Answers (2)

Andreas Wenzel
Andreas Wenzel

Reputation: 24726

If, for example, the user enters the value 3, then the following will happen:

  1. The function main will call countingdown with the parameter 3.
  2. The function countingdown will then compare this parameter 3 with the value 1, and, since that comparison evaluates to false, it will print the number 3 and then call countingdown (i.e. the function will call itself) with the parameter 2.
  3. This second instance of countingdown will now compare this parameter 2 with the value 1, and, since that comparison evaluates to false, it will print the number 2 and call countingdown with the parameter 1.
  4. This third instance of countingdown will now compare the parameter 1 with the value 1, and since that comparison now evaluates to true, it will print the number 1 and will not call itself again.

The recursive function treats the parameter 1 as special, since that is the value at which it should abort. When it encounters this value, it will stop calling itself recursively. If the function did not check for this special value, it would continue counting down forever and would also print 0, -1, -2, etc.

Upvotes: 1

EsmaeelE
EsmaeelE

Reputation: 2658

You can rewrite the recursive function countdown()

Similar to this 

int countingdown(int n){

    if (n == 0){//return exit from recursion
        return;
    }
    else {
        printf("%d\n", n);
        countingdown(n - 1);
    }
}

Suppose we call it with countdown(5). Until we reach to 0 printing n, i.e 5, 4, 3, 2, 1 will continue. In last step that variable n reach to 0 as we express in if section function returns and its execution ends.

[Edit]

Better one

void countingdown(unsigned int n){

    if (n){
        printf("%d\n", n);
        countingdown(n - 1);    
    }

}

Upvotes: 1

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