hello i want to know the difference between pre and post increment. from what i know is that both thing performs addition over a variable .
How ever considering this example


The value is stored in a variable called a. Then the program will show the output as seen below:
The value of a is 10.
................................
The value of ++a is 11.
The value of a is 11.
The value of a++ is 11.
The value of a is 12.
The value of --a is 11.
The value of a is 11.
The value of a-- is 11.
The value of a is 10.

static void Main(string[] args)
{

int a = 10;
Console.WriteLine("a is {0}",a); // 10

int b = ++a;
Console.WriteLine("++a is {0}",b); //11
Console.WriteLine("a is {0}",a);//11

b =a++;
Console.WriteLine("a++ is {0}",b);//11
Console.WriteLine( "a is {0}", a);//12

b = --a;
Console.WriteLine("a-- is {0}",b);
Console.WriteLine("a is {0}",a);//11


b = a--;
Console.WriteLine("a -- is {0}",b );//11
Console.WriteLine("a is {0}",a );//10


why is it that in case b= a++ the increment does not show on the output however it shows for a ?

It is infact the same for a-- aswell.

b=++a; means to take a, plus one and then assign it to b.
b=a++; means take a, assign it to b and then add one to a.

same for --

_________________
www.seuntjie.com

That is the Misconcept , most people think that way for instance consider this example :
tatic 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

Although the increment / decrement is done(in case of post ) but still the output on console is not the incremented or decremented value but the original value that variable holds ?
now my question is why ?

No that is not the misconception.
++a means increment the value and then use it. a++ means use the value then increment it, Wether your using int x =a++; or console.writeline(a++);

_________________
www.seuntjie.com

i just wanted to enlighten u with the correct information now what ever u believe is your choice

a++ known as postfix.
postfix:: First use its expression, then increase the value
++a known as prefix.
prefix:: First increase the value, then use its expression

so: Seuntje is right.

in both the operations (prefix and post fix)the execution is exactly in same order but the only difference is that in postfix ,the result of the operation is the temporary copy of the original value that is why the answer is the same old value in case of a++ ,although the increment has taken place yet we cant see and the proof of it (i have already mentioned)

that is how i look at it i am not compelling u to believe it infact i was myself dubious about this thing that lead me to study about it

@Nagisa dude if you ask a question like this and get a good answer from a qualified programmer ( I have a B.Sc in programming), don't argue.

for all practical purposes my original answer is correct. If you want detailed information on what happens inside, ask how it happens inside a++ and ++a that makes them different, not why the values are different, as in your first post.

_________________
www.seuntjie.com

@seuntjie i am not doubting ur abilities neither do i need an info about ur qualification..
but yes the execution steps in case of a++ and ++a is something that mattersand i tend to seek insight knowledge about all cores and basics
Like i said i am not challenging anyone, i am just sharing my knowledge on the subject