Argthtjtr

This recursive function takes two input arguments, The first (A) is a number and the second (n) is a digit, checks the occurrence of n in A. (A is updated by removing its last digit in each recursion). it seems like the recursion is infinite and the base case (A == 0) is not valid but why.

function counts = countn(A,n)

    if (A == 0)

        counts= 0;

    end

    if (n == mod(A,10))

        disp(A);

        disp(floor(A/10));

        disp(mod(A,10));

        B = floor(A/10);

        counts = countn(B,n) + 1;



    else 

       B = floor(A/10);

       countn(B,n);

    end 

    end

It does not stop because it first evaluates the first if statement if( A == 0) and afterward the if (n == mod(A,10)) where it jumps in the else branch and recursively calls the function again. So it does not stop in the first if statement as you likely expected it to do.

something like this should work:

function counts = countn(A,n)

    if (A == 0)

        counts = 0;

    elseif (n == mod(A,10))

            disp(A);

            disp(floor(A/10));

            disp(mod(A,10));

            B = floor(A/10);

            counts = countn(B,n) + 1;

    else 

       B = floor(A/10);

       counts = countn(B,n);

    end 

end

You also have to update count counts variable in the else branch to avoid the uninitialized use of variables.

Have a look at how to use a debugger manual. Simply click on the line number inside your function and run your code. Use the F10 and F11 keys to evaluate your code line by line. This helps you understand what your program does.