mechanism of calling of merge sort algorithm

Question

1st mergesort function is called till the if condition is satisfied Now according to my knowledge if if-conditon is not satisfied then its body will not execute

I m unable to understand after the unsatisfaction of if-conditon 2nd mergesort function is called.so how it happens? Plz explain in detail

mergesort (int*a,int*b,int low,int high)

{

     int  pivot
     if(low<high)
      {
               pivot=(low+high)/2;
               mergesort(a,b,low,pivot);/*1st*/
               mergesort(a,b,pivot+1,high);/*2nd*/
               merge(a,b,low,pivot,high);
       }
     return ;
}

Answer 1

To follow up on Rohlex32's answer (his answer should get credit for the correct answer).

my question is why mergesort(a,b,pivot+1,high) is called?

As answered by Rohlex32, it's because mergesort(a, b, low, high), uses high as the last index to be sorted, so mergesort(a, b, low, pivot), includes the element at [pivot].

An alternative implementation of mergesort uses a beginning and ending index, where the ending index is 1 greater than the last index. The initial call to mergesort would be

mergesort(a, b, 0, sizeof(a)/sizeof(a[0]));

Such a mergesort function would be a bit different than above:

void mergesort(int *a, int *b, int begin, int end)
{
    int mid = (begin+end)/2;
    if((end - begin) < 2)
        return;
    mergesort(a, b, begin, mid);    // sort from begin to mid-1
    mergesort(a, b, mid, end);      // sort from mid   to end-1
    merge(a, b, begin, mid, end);   // merge the two parts
}