Hello...why this function doesnt work for n=0?-CodePudding

I have to write a recursive function in C to convert a number from decimal to binary. This is the prototype we were given void dec_to_binary(int n)

 
void dec_to_binary(int n)
{
    // base case: if the number is 0, return
    if (n == 0)
    {
        return;
    }
    // recursive case: divide the number by 2 and call the function again
    dec_to_binary(n / 2);
    // print the remainder (which will be 0 or 1)
    printf("%d", n % 2);
}

CodePudding user response：

Your code does not print anything when it is zero because it returns immediately. Perhaps you could check if input is zero before calling the function, like this:

void dec_to_binary_recursive(int n)
{
    // base case: if the number is 0, return
    if (n == 0)
    {
        return;
    }
    // recursive case: divide the number by 2 and call the function again
    dec_to_binary_recursive(n / 2);
    // print the remainder (which will be 0 or 1)
    printf("%d", n % 2);
}

void dec_to_binary(int n)
{
    if (n == 0)
    {
        printf("%d", 0);
    }
    else
    {
        dec_to_binary_recursive(n);
    }
}

Keep in mind that this does not support negative numbers, and there probably a better way

CodePudding user response：

As noted above your current dec_to_binary function immediately returns and does nothing when the input is 0.

You can solve the problem without adding a function,
by having 2 base cases for the recursion: for 0 and 1:

#include <stdio.h>


void dec_to_binary(int n)
{
    if (n == 0)
    {
        printf("0");
        return;
    }
    if (n == 1)
    {
        printf("1");
        return;
    }
    dec_to_binary(n / 2);
    printf("%d", n % 2);
}


int main()
{
    dec_to_binary(0);
    printf("\n");
    dec_to_binary(1);
    printf("\n");
    dec_to_binary(2);
    printf("\n");
    dec_to_binary(7);
    printf("\n");
}

Output:

CodePudding user response：

It's possible to make an way that does not need an wrapper function or a second argument. Does not work for negatives tough

void dec_to_binary(int n) {
    if (n > 1)
        dec_to_binary(n/2);
    printf("%d", n % 2);
}