Can someone explain this output? (C++)

Question

Hello can someone please explain the output of the following C++ code espacially the numbers after the output of the first one 43211223334444

Here is the code:

void rek(int i) { 
if (i > 0){
   cout << i; 
   rek(i-1); 
   for (int k=0; k < i; k++) 
      cout << i; }
}
int main(){
 rek(4);
 return 0;
}

Here is the output:

the output is: 43211223334444

1. clang format this. 2. Learn to use debugger ASAP 3. Did you noticed for loop? — Marek R, Jun 18 '21 at 10:03
By the way! Welcome to SO! Please take a look at ["How to ask a good question"](https://stackoverflow.com/help/how-to-ask) it will help you down the road. Also, please, don't forget to mark an answer as accepted if you consider one as such. — rawrex, Jun 18 '21 at 11:10
[What is a debugger and how can it help me diagnose problems?](https://stackoverflow.com/q/25385173/995714), [How to debug small programs](https://ericlippert.com/2014/03/05/how-to-debug-small-programs/) — phuclv, Jul 11 '21 at 10:18

rawrex · Answer 1 · 2021-06-18T10:23:59.667

All the trouble comes from the fact that your output didn't introduce any separation in values cout << i; prints.

You actually getting the following at first:

4, 3, 2, 1

We got this out of these two statements:

// ...
cout<< i;
rek(i-1);
// ...

Where each call to the rek(i) prints its current i and then calls rek(i-1), which in turn before proceed through the function body has to print its i-1 and then call rek((i-1)-1)... And so on, until it hits the i < 0 case.

The remaining part is brought to you by the output of 1 one time, 2 two times, 3 three times, and 4 four times. All because of the for loop:

for (int k=0; k < i; k++) // print the value of i, i times
      cout << i;

Thus, you essentially have the following:

4
3
2
1 // We hit base case on next rek() call, so no print out of there
1 // Here the for loop comes in
2 2
3 3 3  
4 4 4 4

By the way, please note that the placement of brace(s) in your code is somewhat counterintuitive.

Peter - Reinstate Monica · Accepted Answer · 2021-07-11T09:34:25.440

We can write the dynamic stack development down manually. Below I have pseudo code with an indentation of four per stack level. The commands are on the left, the resulting output, in that order, on the right. I have replaced i with the respective number to make even more transparent what's going on.

rek(4)                                                              
    cout << 4;                                      4
    rek(3)
        cout << 3;                                  3
        rek(2)
            cout << 2;                              2
            rek(1)
                cout << 1;                          1
                rek(0)
                    if(i>0) // fails, rek(0) returns
                rek(1) continues
                for (int k=0; k < 1; k++)
                    cout << 1;                      1
                rek(1) returns
            rek(2) continues
            for (int k=0; k < 2; k++)
                cout << 2;                          2
                cout << 2;                          2
            rek(2) returns
        rek(3) continues
        for (int k=0; k < 3; k++)   
            cout << 3;                              3
            cout << 3;                              3
            cout << 3;                              3
        rek(3) returns
    rek(4) continues
    for (int k=0; k < 4; k++)   
        cout << 4;                                  4
        cout << 4;                                  4
        cout << 4;                                  4
        cout << 4;                                  4
    rek(4) returns
main() continues
return 0;

score 1 · Answer 3 · answered Jun 18 '21 at 10:11

Lets break your output string: "43211223334444" into 2 parts:

Part 1: "4321" This is the result of
```
cout << i; 
rek(i-1); 
```

You are printing i and recursively calling same function by passing the (i-1) as argument and function performs the operation till (i > 0). This prints the numbers from 4 to 1, i.e. "4321"

Part 2 "1223334444" This is the result of code section:
```
for (int k=0; k < i; k++) 
   cout << i;
```

This section gets called in reverse order from number 1 to 4. This code section basically prints the number i, for the i times.

For i=1 it prints: 1
For i=2 it prints: 22
For i=3 it prints: 333
For i=4 it prints: 4444

That makes the string: "1223334444"

Hope this explains you the total output string: "43211223334444"

score -1 · Answer 4 · answered Jun 18 '21 at 10:08

-1

you are doing recursive function and then a for-loop and you're function is going deep in the recursion and then the for loop is working from the deepest point of the recursion that's why you see this output

answered Jun 18 '21 at 10:08

M7mood1996

1

score -1 · Answer 5 · answered Jun 18 '21 at 10:10

You start with going "down in the recursive tree", first you complete printing the numbers from i=4 to i=1 and therefore you print: 4321. After that you go "up the recursive tree" and start printing from i=1 to i=4 using the loop you wrote, therefore the order is now for i=1 you print 1 and for i=2 you print 22 which result in printing 1223334444 at the end of the line.

Can someone explain this output? (C++)

5 Answers5