Why does this recursive function generate this output? - c

Question

I am trying to understand the output of the program printed below. When I look at it, I see that when printnum() is called with an argument of 1, "1" will be printed and then since 1<7, the function will call itself. This process will continue until "6" is printed and then printnum(7) is called. So, now "7" is printed and the if condition is not satisfied, so that code is skipped and we move to the second printf("%d", x) function where "7" is printed out again. There is nothing after the second printf("%d", x), so why doesn't everything end there? What makes the program keep going to print the numbers again in descending order?
#include <stdio.h>
int printnum ( int x )
{
printf("%d", x);
if ( x < 7 )
{
printnum ( x + 1 );
}
printf("%d",x);
}
int main()
{
printnum(1);
}
Output:
12345677654321

Answer 1

printnum(8) is never called, because it isn't true that 7 < 7.
The reason that you get the numbers printed in descending order once x = 7 is reached is, that each recursive call ends, leaving the previous call to continue.
Consider what it does for x = 1:
Print 1
Call recursively with x = 2
Print 1
If we expand this one level more:
Print 1
Print 2
Call recursively with x = 3
Print 2
Print 1
And one more:
Print 1
Print 2
Print 3
Call recursively with x = 4
Print 3
Print 2
Print 1
If you continue this expansion, you can see you get numbers in ascending order before the recursive call, and in descending order after the recursive call.

Answer 2

This happens because the second printf is called after your recursion exits at each level.
As your final recursive call printfs and ends, control transitions to the function that called it - the second-to-last recursive call. This call then exits the scope of the if statement, and calls printf, and then ends - upon which control transitions to the function that called it - the third-to-last recursive call. Repeat until you are inside the call to printnum(1), whose return takes you back to main.

Answer 3

It's recursion. When you enter the function you call printf, then you enter another level in the recursion, then you enter the printnum, so you call the printf with x+1 and so on.
When you arrive at stop condition (x==7), the function goes till the second printf (so 7 will be displayed again).
The function printnum terminates, so the program returns at the level above, then printnum at level 6 can printf again, terminate and return and so on.

Answer 4

Comments in-line!
int printnum ( int x ) x = 1 x = 2 x=6 x=7
{
printf("%d", x); prints 1 prints 2 prints 6 prints 7
if ( x < 7 ) Yes Yes Yes No
printnum ( x + 1 ); calls printnum(2) calls printnum(3) .... calls printnum(7) N/A
printf("%d",x); print 2 and return to print 6 and return prints 7 and returns to the
the caller which is the previous caller to the previous previous caller which is
main() which is printnum(1) caller which is printnum(x=6)
printnum(5)
}
Please check the following to print in ascending and descending order passing the starting value and the limit.. (please note that the error checking is not done!)
#include <stdio.h>
int printnum_123(int x, int limit)
{
printf("%d ", x);
if (x<limit)
printnum_123(x+1, limit);
return;
}
int printnum_321(int x, int limit)
{
if (x<limit)
printnum_321(x+1, limit);
printf("%d ", x);
return;
}
int main(void)
{
printnum_123(1, 10); printf("\n");
printnum_321(1, 10); printf("\n");
return 0;
}
$ ./a.out
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
$