how does recursion method helps in printing array

Question

void printarray(int i) {
    if (i == 0) {
        return;
    } else {
        printarray(i - 1);
    }
    System.out.println("[" + (i - 1) + "]" + values[i - 1]);
}

Its a program that prints the array using recursion. Can anybody tell me how does the above method works??

The debugger can for sure. Make a pen&paper test. Use a separate sheet of paper for each new call to `printarray(...)`. — Turing85, May 23 '17 at 18:03
Attach a debugger in the IDE you are using and check for each step. — Devendra Lattu, May 23 '17 at 18:06
Even simpler, insert one or two print statements to trace control flow and data values. — Prune, May 23 '17 at 18:08
There are many explanations of recursion on the Internet, many explanations on Stack Overflow. Where are you lost? — Prune, May 23 '17 at 18:09
@Prune i have tried to understand it by putting print statements but i am confused in the last line only — arunoday singh, May 23 '17 at 18:18
i have search it but not able to understand it that why i have put it here.don't think that i haven't searched it and directly put up this question — arunoday singh, May 23 '17 at 18:19
@arunodaysingh make a pen&paper test, as I suggested in my first comment. — Turing85, May 23 '17 at 18:24

cyberbrain · Accepted Answer · 2017-05-23T18:18:21.407

1

The method prints an array that has to be an instance or class variable with the name values.

The initial given index is not taken into account.

The method starts at the topmost index and calls itself with an index decremented by one. When it reaches index 0, this is the start of the return-chain. After returning from the recursion call, the value at the current index is printed - after the index itself.

One call is printed under the next. The lines start at index 0 and stop at index i - 1 (the index used for the initial call).

Update: thanks to @Turing85 I fixed my own assumption of the printed indexes.

edited May 23 '17 at 18:18

answered May 23 '17 at 18:11

cyberbrain

3,433
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2

@Turing85 i want to know 1 thing.if i==0 it returns then in the last line the value of i beomes 1.am i right??? – arunoday singh May 23 '17 at 18:25
@arunodaysingh I am not able to follow your thought. A method is immediately aborted, if a `return` is reached. From this point, execution continues one call stack above, (most likely `printarray`). This call frame has its own (old) value of `i` since Java is pass-by-value always. – Turing85 May 23 '17 at 18:28
2

@Turing85 thanks a lot for your help. actually i forgot the call by value.i wasnot able to understand how the last line started with i=1.but with your comments i am able to understand.thanks a lot for your help.. – arunoday singh May 23 '17 at 18:39

score 0 · Answer 2 · answered May 23 '17 at 18:19

0

Example: i = 3 => printarray(3)

else block is active: printarray(2)
else block is active: printarray(1)
else block is active: printarray(0)
if block is active, because i = 0 and return

answered May 23 '17 at 18:19

Patzi0207

13
7

how does recursion method helps in printing array

2 Answers2