Can somebody give me the reason why the integer i is not getting incremented.
Here is the code:

int i=10;

and the output is 10.

Here my doubt is that, i know that i is post incremented. Eventhough i is not incremented at the time of assignment, i should get incremented after that instruction(post incrementation) ie the printf statement should print the incremented value as in the below example

int i=j=10;
printf("%d %d",i,j);

here the output is 10 11. and i kno the reason that i gets the value before the increment and j is incremented(post incremented) after the completion of the instruction.
My question is y the same thing is not happening in the previous example

Waiting for your valuble explanations..

That's because of the way post-increment works.. it's implemneted something like this:

int operator++(int a) //post increment (i++)
     int tmp = a ; //where "a" is the variable you're incrementing
     a = a + 1 ;
     return tmp ;

int operator++(void) //pre increment (++i)
     //this is only an example..
     //where "this" is the pointer to the variable you're incrementing
     *this = *this + 1 ;
     return *this ;

int main()
   int i = 10 ;
   i = i++ ;
   //when post increment operator is called it increments i
   //but >>returns the original<< value viz 10.
   //AFTER the increment assignment happens, so i is reset to 10.
   printf("%d\n", i) ; //will print 10

   i = ++i ;
   //when pre increment operator is called it increments i
   //and >>returns the incremented<< value viz 11.
   printf("%d\n", i) ; //will print 11
    return 0 ;

When you use a variable twice in an expression and one of those instances is a pre- or post- increment, the behavior is undefined (and then depends on how it's implemented by your compiler).

commented: Good link :) - Salem +6

++i and i++ are all 11 in .net

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