Need help with recursive

[qvyhnl]

Can someone show me how to write a recursive function that count how many time the number 7 occur in an integer. Thanks

[mrnutty]

Yes I can. But instead of giving you the answer. I am going to help and guide you through this long and tough journey to find the number of magical sevens in a given integer.

So to start, do you have any ideas? Does not have to be in code, just basic ideas?

[qvyhnl]

I tried to get the last digit my using %, and then remove it with /
but i think my if statement is wrong, as well as setting up that base case.

int count7(int n)
{
    if(n % 10 != 7) return 0;
    if(n == 7 || n % 10 == 7) return 1 + count7(n/10);
}

[mrnutty]

Your attempt is close. You need to change line 3. Also line 4 can be reduced to simply if(n % 10 == 7) return 1 + count7(n/10) Line 3 is going to be your base case. When should you stop in ALL cases? For example in you base case if n = 71, then (71 % 10 != 7) == true, hence your function stops and returns 0.

Base cases is a case where the input n is the minimal number possible for the recursion to work. So what range of value passed in should the function automatically just stop?

[qvyhnl]

it seems like my code stop working whenever the next number is not 7...
I am thinking that the if statement should stop when I get to the last number and it is not equal to 7
but I am failing to make it work.
i tried if(n/10 == 0 && n != 7) return 0;

[mrnutty]

Not quite, you need to check if n < 7, if so then stop and return 0 else you should check each digit for a seven.

[qvyhnl]

i tried that also, but it failed even more

int count7(int n)
{
    if(n < 7) return 0;
    if (n % 10 == 7) return 1 + count7(n/10);
}

[mrnutty]

Yes, your not quite done yet. Your line 4 is not quite yet correct. Here is what you have to do:

check if the last digit is a 7, if so then return 1 + count(n/10) if not then return 0 + count(n/10)