What is the output of the following code? And please why :)

Question

Thank you so much for taking time to answer my question first of all.

What is the output of the following code? And why please ?

int loopy (int i) {
if (i==0) {
printf("%i", i);
}
else{
loopy(i - 1);
printf("%d", i);
} }
loopy(5);

Answer 1

visph Jayakrishna🇮🇳 I am sorry I know that my question seems elementary, but since the recursive call is before the print statement. why doesn’t it only print 0 since this is when the recursion stops. why would it print the other numbers ?

Answer 2

that's a recursive function: a function wich call itself before printing the received argument until modified argument reach a base case (case when the recursion stop)..

does it make sense for the output you get by running it in a code playground?

Answer 3

012345
Its recursive function...
Recursive calls are
loopy(5) initial, it calls loopy(4) and at last after this call returns prints i=5
loopy(4) , it calls loopy(3) and at last after this call returns prints i=4
loopy(3), it calls loopy(2) and at last after this call returns prints i=3
loopy(2) , it calls loopy(1) and at last after this call returns prints i=2
loopy(1) , it calls loopy(0) and at last after this call returns prints i=1
loopy(0) prints i=0 at first and returns to loopy(1)..

Answer 4

visph I got (012345) but shouldn’t it be (543210)? i just don’t understand why :)

Answer 5

if you have a linux desktop/laptop you could run the code line by line by using GDB (GNU Debugger) to better figure how it works:

https://www.gnu.org/software/gdb/

Answer 6

no, because the recursive call occurs before the print statement ;)

Answer 7

because after printing zero it return just after the call, print 1, then recursively return and print all nums from 2 to 5 ^^

Answer 8

Because when it calls loopy(5-1) i=5, not printed until loopy(4) calls completes and returns but it doesn't take any value back.. (because loopy(4) is entirely different call from loopy(5)) so i value still 5 for loopy(5).

Similarly loopy(3), loopy(2), loopy(1).
loopy(0) not make any calls recursively...
Hope you understand it.

Answer 9

loopy(5) = loopy(4) then print(5)
= Loopy(3) then print(4) then print(5)
= Etc.

See ? Loopy(4) is replaced by one iteration of the loop, leaving a print() while the loop continues above it.