post-pre increament

To understand how pre and post increment work, we need an example different from the one you wrote; look at this:

int i = 0;
int j = i++;
int k = ++i;
printf("i = %d\nj = %d\nk = %d\n", i, j, k);

This will produce an output like:

i = 2
j = 0
k = 2

The statement int j = i++; means: first assign to j the value of i and after increment i (it is the same than int j = i; i = i + 1;)
The statement int k = ++i; means: first increment i and after assign to k the value of i (it is the same than i = i + 1; int k = i;)

The part after the second semicolon in a for line -
will be processed after the first loop primary :)

So you can feel the difference of ++x & x++ (--x & x--)
for example by the following lines:

{
  // one result
  int i = 5;
  do {
    // after-call decrement
    printf("%d\n", i--);
  } while(i);
  // another result
  printf("\n");
  i = 5;
  do {
    // pre-call decrement
    printf("%d\n", --i); 
  } while(i);
}

To add a bit of drama to the answer, look at these lines of code.

int ID = 1000;
int IX = 1000;
int GetNextID() {return ID++; }
int GetNextIX() {return ++IX; }

The first one returns 1000 and the second 1001. But if you check, both ID and IX will be 1001 ! Thats because, for ID, increment is done after the return while it is done before for IX.

You may ask how the heck can a code execute after return but know that return translates into a couple or more lines of assembly code. When debugging, watch these variables, step into each line in dis-assembly window and observe.

I hope now you'll never forget how these work. Ever.