使用C ++中的mergesort实现的特殊行为，排序结果不应到位

Question

In the implementation below of mergesort, the input pointer inputIntArray is somehow kept from being destroyed and winds up pointing to the sorted values pointed by sortedArray . As it is, I thought inputIntArray would not point to anything given the call delete[] intArray; just before the last merge. Could anyone please shed some light on this? Thank you.

#include <iostream>
using namespace std;

int* getSubArray (int*,int,int);
int* merge(int*,int*,int,int);

int* mergeSort ( int* intArray , int n ) {
    if (n==1){
        return intArray;
    }
    else {
        int m = n/2;
        int* leftArray = mergeSort ( getSubArray (intArray,0,m-1) , m );
        int* rightArray = mergeSort ( getSubArray (intArray,m,n) , n-m );
        delete[] intArray;
        return merge(leftArray,rightArray,m,n);
    }
}

int* getSubArray ( int* intArray , int start , int end ){
    int n = end - start + 1;
    int * subIntArray = new int [ n ];
    for (int i=0;i<n;i++){
        subIntArray[i]=intArray[i+start];
    }
    return subIntArray;
}

int* merge(int* leftArray, int* rightArray, int m, int n){
    int* intArray = new int[n];
    int i=0 , j=0;
    for (int k=0;k<n;k++){
        if ( i == m && j == n ) {
            break;
        }
        if (leftArray[i]<=rightArray[j] && i < m){
            intArray[k]=leftArray[i++];
        }
        else {
            intArray[k]=rightArray[j++];
        }
    }
    delete[] leftArray;
    delete[] rightArray;
    return intArray;
}

int main () {

    int* inputIntArray = new int[6]{1,5,6,2,3,7};
    for (int i=0;i<6;i++){
        cout << inputIntArray[i];
    }
    cout << endl;

    int* sortedArray = mergeSort ( inputIntArray , 6 );
    for (int i=0;i<6;i++){
        cout << inputIntArray[i];
    }
    cout << endl;
    for (int i=0;i<6;i++){
        cout << sortedArray[i];
    }
    cout << endl;
    delete[] sortedArray;
}

Here is the output:

156237
123567
123567

Answer 1

You have this call:

int* sortedArray = mergeSort(inputIntArray, 6);

In the process of executing that code, you do:

delete[] inputIntArray;
sortedArray = new int[6];

It is certainly possible that sortedArray gets the same address as the just deleted inputIntArray .