Question

I know that values sent to a function are by default passed as value and the method receives a copy of the variable. I know that when a variable is passed by reference the method can change the value of the variables from it was called. That being said, can someone help explain what’s going on in these simple illustrations? thanks in advance. Are expressions passed by reference, I guess?

using System;
class Program
{
    static void Main(string[] args)
    {
        int x = 2;
        int y = 20;
        Console.WriteLine(Add(x, y));
    }

    static int Add(int x, int y)
    {
        int ans = x + y;
         x = 20;
         y = 40;
       // return x+y;
         return ans;
        //returns 22
    }
}

and then

using System;
class Program
{
    static void Main(string[] args)
    {
        int x = 2;
        int y = 20;
        Console.WriteLine(Add(x, y));
    }

    static int Add(int x, int y)
    {
        int ans = x + y;
         x = 20;
         y = 40;
       return x+y;
      //   return ans;
        //returns 60
    }
}

Answer 1

Both of these illustrate pass by value, and in general, value type semantics.

In your first example, you are saying:

int ans = x + y;

This evaluates x and y at that time and adds them together to store in ans, setting x and y to new values later does not affect the value of ans.

Consider it this way:

static int Add(int x, int y)
{
    int ans = x + y;  // evaluates ans = x + y = 2 + 20 = 22
     x = 20;          // here ans = 22, x = 20, y = 20
     y = 40;          // here ans = 22, x = 20, y = 40
     return ans;      // returns ans which is still 22 since x & y are independent
}

In your second example, you are delaying the addition till after you set the new values, thus the new values of x and y are used in the computation of

return x + y;

So, in essence, you get:

static int Add(int x, int y)
{
    int ans = x + y;  // ans = 22, x = 2, y = 20
     x = 20;          // ans = 22, x = 20, y = 20
     y = 40;          // ans = 22, x = 20, y = 40
   return x+y;        // evaluates x + y = 60
}

I think what may be confusing you is that:

ans = x + y;

Is not a function, it’s an expression which is evaluated and returned, changing x or y after this statement executes does not affect ans again.

The key point to remember, again, is that ans = x + y; evaluates x and y at the time that statement is executed. Further changes to x and y don’t come into play here.