> Also struggling a bit with
> result *= y++;
> I interpreted that as:
> #local Result = Result * (Y+1);
> (line 102)

The whole statement, when decomposed, can be interpreted in C/C++ as

      old_y = y ;
      y = y + 1 ;
      result = result * old_y ;

'y++' is interpreted as get the value of y, increment y by one then return the
value of y prior to incrementing it.

      #local Old_Y = Y ;
      #local Y = Y + 1 ;
      #local Result = Result * Old_Y ;

In your interpretation, Y will never be incremented ...

Post a reply to this message