pre post increament?

[vishalgoswami]

#include<stdlib.h>
#include<stdio.h>
int main()
{
int i=5;
printf("%d\n",i);
printf("%d\n",i++);
printf("%d\n",++i);
printf("%d - %d - %d\n",i,i++,++i);
}

/*how output is like that?
5
5
7
9 - 8 - 9
*/

[Meetee]

It is because

printf("%d\n",i ); 5, as it is apparent
printf("%d\n",i ++); 5, as it will first print and then increment. After this line execution, i = 6
printf("%d\n",+ +i); 7, as first increment and make it 7 and then print
printf("%d - %d - %d\n",i,i++,++i ); 9 - 8 - 9, I am not very sure, but compiler executes this in reverse order.

[weaknessforcats]

printf("%d - %d - %d\n",i,i++,++i); 9 - 8 - 9, I am not very sure, but compiler

This code will produce indeterminate results. While the comma operator executes the actions in order, there is no provision in the standard as to when the variable will be changed. It just has to be changed before leaving the statement. Therefore, the value in the variable is indeterminate when the change is applied.

Do not modify the same variable more than once in the same statement.

[Shubhangi24]

It is mostly environment dependant. If you will do the same program on TC(Turbo C) it will give you the answer as:
/*
5
5
7
9 - 8 - 8
*/