CODEWITHSUNDEEP
cfunctionrecursiondefinitioncollatz
Jonathan Jerai
Jonathan Jerai

Reputation: 55

main function does not call collatzSequencer function

I am currently working on a project for college, first year computing class. With that being said, I am not asking for the answer, but I am asking more for some advice. In order to begin the project, I have decided to create a function called collatzSequencer that takes one argument type int.

Here is my function prototype: 

int collatzSequencer(int);

Here is my function definition: 

int collatzSequencer(int n) {
    int stepCounter = 0;
    if (n % 2 == 0) {
        stepCounter += 1;
        collatzSequencer(n / 2);
    }
    else if (n % 2 == 1) {
        stepCounter += 1;
        collatzSequencer((3 * n) + 1);
    }
    else if (n == 1) {
        printf("%d\n", n);
        printf("%d\n", stepCounter);
        return 0;

Here is where I call the function within my main function: 

int main(int argc, char* argv[]) {
    int num = 5;
    collatzSequencer(num);
    return 0;
}

When I run my program, nothing happens and I exit with code 0. I have tried debugging my program, and I see that for some reason my IDE doesn't even run the collatzSequencer function when it is called. Although I am a beginner, I feel like I have enough knowledge to be able to find problems within only 48 lines of code, however I cannot find the issue here. Anyone have any ideas?

Upvotes: 1

Views: 45

Answers (2)

Vlad from Moscow
Vlad from Moscow

Reputation: 311068

The function has undefined behavior because it does not return anything when n % 2 is equal 0 or 1 and the last else-if statement is not reached by the function control.

Moreover the variable stepCounter can have the maximum value of 1 because it is a local variable of the function that in each recursive call of the function is initialized by zero.

int stepCounter = 0;

The function can be defined the following way as it is shown in the demonstrative program

#include <stdio.h>

size_t collatzSequencer( unsigned int n ) 
{
    return  n < 2 ? 0 : 1 + collatzSequencer( n % 2 == 0 ? n / 2 : 3 * n + 1 ); 
}

int main(void) 
{
    printf( "%zu\n", collatzSequencer( 27 ) );

    return 0;
}

Its output is

111

as it is written in https://en.wikipedia.org/wiki/Collatz_conjecture

Upvotes: 0

abelenky
abelenky

Reputation: 64710

You are checking for three cases:

n % 2 == 0 (eg. n is Even)
n % 2 == 1 (eg. n is Odd)
n == 1  (eg. n is exactly ONE; this is also the only if-statement with a return)

Except: value 1 is an ODD number, which is captured by the second case.

Your code will never reach the n==1 case, because when n is 1, it will always get captured by the Odd-value if-statement first.

Upvotes: 3

Related Questions