The {0} in C#

[Alien]

Hi,

I am learning C# in csharp station website but theres one part I am little confused and the tutorial doesnt offer any explanation.

Here is the code:

using System;
 
class Unary
{
    public static void Main()
    {
        int unary = 0;
        int preIncrement;
        int preDecrement;
        int postIncrement;
        int postDecrement;
        int positive;
        int negative;
        sbyte bitNot;
        bool logNot;
 
        preIncrement = ++unary;
        Console.WriteLine("pre-Increment: {0}", preIncrement);
 
 

preIncrement should be 1 when console.WriteLine is executed but comes up 0 when I run it.

My understanding is that {0} refers to the 1st Argument since unary which is set to 0 is incremennted before copied over to preIncrement, shouldnt the output be 1 instead of 0?

Cheers

[tlhintoq]

I would agree with that.
Is the code you posted copied from your project, or from the site tutorial?

[r035198x]

Post the full code you ran.

[PRR]

int i=0;
int j=++i;
// This basically means increment i and then assign to j
int k=i++;
//This means assign to k old value of i and then increment i
 

Run this and check for yourself

int i = 0;
            Console.WriteLine("Value of i={0}",i);
            int j=++i;
            Console.WriteLine("----");
            Console.WriteLine("Value of i={0}",i);
            Console.WriteLine("Value of j={0}",j);
            i = 0;
            int k =i++;
            Console.WriteLine("----");
            Console.WriteLine("Value of i={0}", i);
            Console.WriteLine("Value of k={0}", k);
 

[tlhintoq]

int k=i++;
//This means assign to k old value of i and then increment i

No, it doesn't.

Math is not done simplisticly from left to right. The order of operation that we all learned in grade school extends to coding as well.

• Multiplication/division from left to right
• Addition/subtraction from left to right
• Evaluate all of the right side before assigning to the left side of a equal sign

What would the value of A equal here?
a = 3 + 4 * 6
If you do it just left to right you get 3 + 7 = 7 times 6 = 42
If you follow the rules where multiplication is done before addition you get
4 * 6 = 24 + 3 = 27

Everything on the right side of an = is evaluated before it is assigned to the variable on the left side. So in simple formulas like
i = k++;
i = ++k;
there is no difference. The right side is evaluated *THEN* assigned to i

Where you would see a difference is in a comparison because that does take place in simple left to right.

if ( k < i++)
if (k < ++i)

[Markus]

@tlhintoq
Yes, it does...

Mark (must be missing something completely obvious).

P.S.

$x = 1;
$y = $x++;
// prints: x: 2, y: 1
printf("x: %s, y: %s", $x, $y);
 

[cloud255]

While in my experience with C# the result will be:

int x = 0, y = 0;
y = x++;
//x ==1 y == 0
 
----------------------------------------
 
int x = 0, y = 0;
y = ++x;
//x ==1 y == 1
 
 

But this does not explain why the OP gets the stated result unless perhaps the compiler the OP is using does not comply with the C# specifications...

[tlhintoq]

Yes, it does...

OK. I stand corrected. I guess it helps to test things empirically. Using the code you provided I put a breakpoint on your first line, then walked through the code looking at the values after each line of code executes. It seems to behave in the simple left-to-right evaluation that you described. Do you get different results than those shown in the following images?

[Markus]

Correct - that's how I see it happen in PHP and Java (and I've read that's how it works in C).

Mark.

[tlhintoq]

@Markus
And as the screenshots show how it works in C#.
So... what is the problem again?