Function output(#bbsr)

#1

I am not able to understand how the output is coming like this. Can you help me out by a dry run.

#2

HI @nayakashutosh99, what is this code for. like what is the question ?

#3

It’s my doubt .
I don’t have question. I saw in a book and started trying it but my output was coming different. That’s why I asked.

#4

According to me, the code will be implemented as follows :
line 14
line 7 : for value 3
( 3-1) = 2
line 7 : for value 2
(2-1)=1
line 7 : for value 1
(1-1)=0
since n is not >0 for this case, there will be returning from call stack, thus line 8 will be implemented thereafter
line 8 : cout <<“1”
line 9 : (1-1)=0

since again n is not >0 for this case, thus we have returning again, this time value will be 2
line 8 : cout <<“2”
line 9 : (2-1)=1
line 8 : cout<<“1”
line 9 : (1-1)=0;

now again n is not >0, for this case, we have returning again, this time value will be 3
line 8 : cout<<“3”
line 9 : (3-1)=2
(2-1)=1
(1-1)=0
cout<<“1”
cout<<“2”
then again for 2, we have (2-1)=1
thus cout<<“1”

try to understand through the above, if not i will send the flow of call stack…

#5

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