0 votes 0 votes himgta asked Mar 10, 2018 himgta 559 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Sukannya commented Mar 10, 2018 reply Follow Share If it had been pre-increment, then what you are saying would have been true. Steps involved in both the types are shown: (fun--) Step1: fun(n); Step2:n=n-1; Because of the 1st step, function can never be called for the decremented value of n But, fun(--n) Step1:n=n-1; Step2:fun(n); Can you see the difference now... 0 votes 0 votes himgta commented Mar 11, 2018 reply Follow Share thanks.....got it now! 0 votes 0 votes Shiva Sagar Rao commented Apr 13, 2021 reply Follow Share Similar question: https://gateoverflow.in/183214/c-prog 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 5 4 3 2 1 Deeksha Sisodiya answered Mar 10, 2018 Deeksha Sisodiya comment Share Follow See 1 comment See all 1 1 comment reply Sukannya commented Mar 10, 2018 reply Follow Share @Deeksha, this is post-decrement,hence would give infinite loop, see the comments below the other answer 0 votes 0 votes Please log in or register to add a comment.