Pre- & Post Increment in C#

Schweder picture Schweder · Dec 20, 2011 · Viewed 30.3k times · Source

I am a little confused about how the C# compiler handles pre- and post increments and decrements.

When I code the following:

int x = 4;
x = x++ + ++x;

x will have the value 10 afterwards. I think this is because the pre-increment sets x to 5, which makes it 5+5 which evaluates to 10. Then the post-increment will update x to 6, but this value will not be used because then 10 will be assigned to x.

But when I code:

int x = 4;
x = x-- - --x;

then x will be 2 afterwards. Can anyone explain why this is the case?

Answer

Sebastian Piu picture Sebastian Piu · Dec 20, 2011

x-- will be 4, but will be 3 at the moment of --x, so it will end being 2, then you'll have

x = 4 - 2

btw, your first case will be x = 4 + 6

Here is a small example that will print out the values for each part, maybe this way you'll understand it better:

static void Main(string[] args)
{
    int x = 4;
    Console.WriteLine("x++: {0}", x++); //after this statement x = 5
    Console.WriteLine("++x: {0}", ++x); 

    int y = 4;
    Console.WriteLine("y--: {0}", y--); //after this statement y = 3
    Console.WriteLine("--y: {0}", --y);

    Console.ReadKey();
}

this prints out

x++: 4
++x: 6
y--: 4
--y: 2