why preincerment operator is faster than postincrement operator?

Answer 1

becuase number of assembly instructions are less (however nowadays compiler are smart enough to optimize both ++i and i++ )

Answer 2

Because the language can load and add 1 in one step, then use the result. To increment later, the value just used must be loaded again, the addition of 1 done and the value put back in a register for use later. Simply said, in any system, there are 1 or 2 less steps possible in pre vs post increment

Answer 3

preincremental operator is faster than postincremental operator only in those cases where operation is done on the variable after incrementation.
Consider the example a[x++]........what happens here is that if x=7, then a[7] is to be done and then x is to be loaded again ad then incremented.{ x is loaded, a[;....] is performed, x is again loaded and then incremented.]

While in case of a[++x], the value of x is loaded, incremented and then a[....] is performed.

So the number of assembly langusae steps required is less for a preincremental than taht of a post incremental.