don't understand the order of this code lines.

Question

Hi

could somebody explain me how this code works?
Suppose number = 123. This code gives '1' '2' and '3'.

I came to the conclusion that the last line (Console.Write(" {0} ", n % 10);) which is 23, is executed once when the return is reached. But how is that possible because the return is in another block of code? What is teh logical order of this code?
Thanks

static void Main()
{
Console.Write(" Give a number: ");
int num = Convert.ToInt32(Console.ReadLine());
Console.Write(" The digit of {0} are: ", num);
cijfers(num);
Console.Write("\n\n");
}

static void cijfers(int n)
{
    if (n < 10)            
    {
        Console.Write("  {0}  ", n);
        return;
    }
    cijfers(n / 10); 
    Console.Write("  {0} ", n % 10);
}

Answer 1

Why with number 123 are the two last lines executed after 'return' and not with number '1'

This is a recursive method. Don't think of it as one method. When the number is 123, these are three methods.

After the return is executed, as we know, the code after the return statement in this method will not be executed, but this will not affect other methods.

In this example, return is executed only once in the last method. In the first two methods, the code does not enter the if block.

It might be easier to understand to translate it into a normal method, it's like(when number is 123):

        static void cijfers(int n)  
        {  
            cijfers1(n / 10);  
            Console.Write("  {0} ", n % 10);  
        }  
        static void cijfers1(int n)  
        {  
            cijfers2(n / 10);  
            Console.Write("  {0} ", n % 10);  
        }  
        static void cijfers2(int n)  
        {  
            Console.Write("  {0}  ", n);  
            return;  
        }  

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