Can't understand this code?

Question

Can anyone help me in understanding the following code:-

int r, countIt(int n) {
    while (r += "            2  "[n % 10] & 3, n /= 10);
    return r;
}

I found this code in one of the challenges of codefights.com, https://codefights.com/challenge/v5Zg8trjoun3PTxrZ/solutions/Aj3ppbhSShixt4nBi

This is a solution for counting number of holes in a number.
e.g.

1111 = 0  
0000 = 4  
1234 = 0  
8888 = 8   

I am not able to understand the following things:
1. Logic of this code
2. comma (,) operator used in return data type of the function
3. Use of []operator after string.
4. And actually the whole code.

Answer 1

I looked up the link you provided. After carefully observing the code i came to following conclusion.

int r, countIt(int n) {.....}

is equivalent to writing as

int r;
int countIt(int n){.....}

now for

while (r += "            2  "[n % 10] & 3, n /= 10);

is equivalent to:

do{
    r += "           2  "[n % 10] & 3;
    n/=10;
}while(n);

Now comes the logical part of the code

r += "           2  "[n % 10] & 3;

let me give you some basics.

1. In c++

cout<<"abcde"[2];

will give you output

c

now if you watch carefully the code in link which you provided its something like this:

r += "           2  "[n % 10] & 3;

is nothing but

r += "TAB,SPACE,SPACE,SPACE,SPACE,SPACE,TAB,SPACE,2,TAB"[n % 10] & 3;

Now its time to explain how this code is calculating number of holes. The ASCII value of TAB is 9 whose binary equivalent is 1001. The ASCII value of SPACE is 32 whose binary equivalent is 100000.

so bit wise anding TAB with 3 will result

            1001 & 0011 = 0001    which is 1

bit wise anding SPACE with 3 will result

            100000 & 000011 = 000000   which is 0

replacing TABs with 1 and SPACEs with 0 hence this concludes as writing

do{
    r += "1000001021"[n % 10] & 3;
    n/=10;
}while(n);

n % 10 is the low-order decimal digit of n. We use that as an index into a string literal, which contains information about how many holes is there in that low-order decimal digit then add it to result r.