Confused about C# function output

Question

When I worked through this example in my head I got an output of 4 8 3 . When I run the function however I get the output 4 8 6 , I understand how to get the 4 and the 8 but I can't understand how y = 6 . Shouldn't y = 3 ? a1 runs which results in y +=1 so y = 1 then a2 runs which results in y+=2 so y = 3 .

void Main() {
    int y = 0;
    Func<int,bool> even = (n) => { return n%2 == 0; };
    Func<int,int> dub = (n) => { y += 2; return n + n; };
    Func<int,int> succ = (n) => { y += 1; return n + 1; };

    Func<bool, int, int, int> if1 = (c, t, f) => c? t: f;
    y = 0;
    var a1 = if1(even(3), dub(3), succ(3));
    var a2 = if1(even(4), dub(4), succ(4));

    Console.WriteLine("{0} {1} {2}", a1, a2, y);
}

Answer 1

Eventhough you have a conditional expression in if1 that will only use t or f , the values send into if1 are always calculated before the call.

To only calculate the values when needed, you would send in delegates to the function, not values:

Func<bool, Func<int>, Func<int>, int> if1 = (c, t, f) => c ? t() : f();
y = 0;
var a1 = if1(even(3), () => dub(3), () => succ(3));
var a2 = if1(even(4), () => dub(4), () => succ(4));

Answer 2

What you have to realize here is that the two calls to if1 are passed the return values of dub and succ (because you actually call them).

var a1 = if1(even(3), dub(3), succ(3));
var a2 = if1(even(4), dub(4), succ(4));

That means that regardless of whether or not the return value is used (as determined by if1 ) the methods have run, and modified y . That is why it is 6, as both dub and succ were called twice by the program, and 2+2+1+1=6