## digitalmars.D.learn - Making sense of recursion

• zbr (23/23) Jun 25 2018 Hi, this question is not specifically D related but I'll just ask
• ag0aep6g (17/41) Jun 25 2018 You seem to think that a recursive call takes over completely, and that
• Colin (7/10) Jun 25 2018 Your mistake is in your visualization :-)
• Timoses (5/28) Jun 26 2018 It's a stack (https://en.wikipedia.org/wiki/Call_stack).
```Hi, this question is not specifically D related but I'll just ask
anyway. Consider the following snippet:

void mergeSort(int[] arr, int l, int r)
{
if (l < r)                       // 1
{
int m = l+(r-l)/2;            // 2
mergeSort(arr, l, m);         // 3
mergeSort(arr, m+1, r);       // 4
merge(arr, l, m, r);          // 5
}                                // 6
}                                   // 7

mergeSort(arr, 0, 4);

When I see this, I visualize the recursion to perform this way:

mergeSort(arr, 0, 4):
0 < 4 ? true: mergeSort(0, 2):
0 < 2 ? true: mergeSort(0, 1):
0 < 1 ? true: mergeSort(0, 0):
0 < 0 ? false: //reach the end of mergeSort /
reach line 6 and then 7

I don't see the computer ever reaching line 4 and 5? Obviously
I'm wrong but where is my mistake?

Thanks.
```
Jun 25 2018
```On 06/25/2018 07:45 PM, zbr wrote:

void mergeSort(int[] arr, int l, int r)
{
if (l < r)                       // 1
{
int m = l+(r-l)/2;            // 2
mergeSort(arr, l, m);         // 3
mergeSort(arr, m+1, r);       // 4
merge(arr, l, m, r);          // 5
}                                // 6
}                                   // 7

mergeSort(arr, 0, 4);

When I see this, I visualize the recursion to perform this way:

mergeSort(arr, 0, 4):
0 < 4 ? true: mergeSort(0, 2):
0 < 2 ? true: mergeSort(0, 1):
0 < 1 ? true: mergeSort(0, 0):
0 < 0 ? false: //reach the end of mergeSort /
reach
line 6 and then 7

I don't see the computer ever reaching line 4 and 5? Obviously I'm wrong
but where is my mistake?

You seem to think that a recursive call takes over completely, and that
the caller ceases to exist. That's not so. mergeSort does call "itself",
but that means there's two active calls now. And when it calls "itself"
again, there's three. And so on. When an inner call returns, the outer
one resumes with the next line as usual.

It's not just a list of recursive calls, it's a tree:

mergeSort(0, 3)
mergeSort(0, 1) // line 3
mergeSort(0, 0) // line 3
mergeSort(1, 1) // line 4
merge // line 5
mergeSort(2, 3) // line 4
mergesort(2, 2) // line 3
mergesort(3, 3) // line 4
merge // line 5
merge // line 5
```
Jun 25 2018
```On Monday, 25 June 2018 at 17:45:01 UTC, zbr wrote:
Hi, this question is not specifically D related but I'll just
ask anyway. Consider the following snippet:

[...]

But... more like:

0 < 4 ? true : mergeSort(0,2) && mergeSort(3, 4)
And so on.

I.e, the it's not either or to run the second mergeSort, they
both happen.
```
Jun 25 2018    Timoses <timosesu gmail.com> writes:
```On Monday, 25 June 2018 at 17:45:01 UTC, zbr wrote:
Hi, this question is not specifically D related but I'll just
ask anyway. Consider the following snippet:

void mergeSort(int[] arr, int l, int r)
{
if (l < r)                       // 1
{
int m = l+(r-l)/2;            // 2
mergeSort(arr, l, m);         // 3
mergeSort(arr, m+1, r);       // 4
merge(arr, l, m, r);          // 5
}                                // 6
}                                   // 7

mergeSort(arr, 0, 4);

When I see this, I visualize the recursion to perform this way:

mergeSort(arr, 0, 4):
0 < 4 ? true: mergeSort(0, 2):
0 < 2 ? true: mergeSort(0, 1):
0 < 1 ? true: mergeSort(0, 0):
0 < 0 ? false: //reach the end of mergeSort /
reach line 6 and then 7

I don't see the computer ever reaching line 4 and 5? Obviously
I'm wrong but where is my mistake?

Thanks.

It's a stack (https://en.wikipedia.org/wiki/Call_stack).

When the program calls a function it is pushed onto the stack. If
that function returns it pops from the stack and the previous
function gets to continue execution from where it stopped before.
```
Jun 26 2018