Generic Segment Tree implementation using C++ Templates - c++

Question

I am trying to make a generic Segment Tree Class for updates and range queries.
Instead of assuming that the elements would just be integers and the operation to be done over a range of elements would be their sum or product, i would want the user to provide the type T of the element and a function, which i named compose.
This function takes in two parameters of type T and returns a value of the same type T. This return value is the result when that desired operation is performed over range of 2 elements which i can use to perform that same operation on a range of any number of elements.
The class is as follows:
#include <functional>
template<class T>
class SegmentTree {
public:
class binary_function_unitype: public std::binary_function<T,T,T> {
public:
virtual T operator() (T arg1, T arg2) {};
};
private:
class Node {
public:
T value;
int seg_start, seg_end;
Node* left;
Node* right;
Node (T value, int seg_start, int seg_end, Node* left=0, Node* right=0) {
this->value = value;
this->seg_start = seg_start;
this->seg_end = seg_end;
this->left = left;
this->right = right;
}
};
// Not expecting the compose function to be robust enough.
T composeUtil (T arg1, T arg2) {
if (arg1!=0 && arg2!=0)
return compose(arg1,arg2);
else if (arg1!=0)
return arg1;
else if (arg2!=0)
return arg2;
}
// Creating the Segment Tree.
Node* createTree (T leaves[], int start, int end) {
// base case - leaf of tree.
if (start==end)
return new Node(leaves[start],start,start,0,0);
// general case.
int mid = start + (end-start)/2;
Node* left = createTree(leaves,start,mid);
Node* right = createTree(leaves,mid+1,end);
T retValue = composeUtil(left->value,right->value);
return new Node(retValue,start,end,left,right);
}
// Range Query helper.
T queryUtil (Node* root, int start, int end) {
int seg_start = root->seg_start, seg_end = root->seg_end;
if (seg_start>end || seg_end<start)
return 0;
else if (seg_start>=start && seg_end<=end)
return root->value;
else
return compose( queryUtil(root->left,start,end), queryUtil(root->right,start,end));
}
// Helper function for Updating the Segment Tree.
void updateUtil (Node* root, int position, T updatedValue) {
int seg_start = root->seg_start, seg_end = root->seg_end;
if(seg_start>position || seg_end<position)
return;
else if(seg_start==seg_end)
root->value = updatedValue;
else
root->value = composeUtil(root->left->value,root->right->value);
}
// Freeing the memory allocated to the Segment Tree.
void destroyTree(Node* root) {
if (root->left!=0)
destroyTree(root->left);
if (root->right!=0)
destroyTree(root->right);
delete root;
}
Node* root;
binary_function_unitype compose;
public:
SegmentTree (T leaves[], binary_function_unitype compose, int start, int end) {
this->compose = compose;
this->root = createTree(leaves, start, end);
}
T query (int start, int end) {
return queryUtil(root, start, end);
}
void update (int position, T updatedValue) {
updateUtil(root, position, updatedValue);
}
~SegmentTree () {
destroyTree(root);
}
};
When I tried to use this class, it turns out that the compose function, which I took in as a paramater is not being used, on the contrary the one from the class binary_function_unitype is being used.
I expected that the function definition from the user would override the one in class binary_function_unitype and my work would be done. But that did not happen. The program using this class is as follows:
#include <iostream>
#include "SegmentTree.h"
using namespace std;
class Compose: public SegmentTree<int>::binary_function_unitype {
public:
int operator() (int arg1, int arg2) {
return arg1+arg2;
}
};
int main()
{
int num;
cin>>num;
int arr[num];
for(int i=0;i<num;i++)
cin>>arr[i];
Compose compose;
SegmentTree<int> segTree(arr, compose, 0, num-1);
int s,e;
cin>>s>>e;
cout<<segTree.query(s-1,e-1);
return 0;
}
Can somebody tell me whats the flaw in my approach or if I misunderstood some basic concept about using inheritance or templates in C++ ?
Thanks.

Answer 1

The constructor takes a binary_function_unitype by value, so it will slice.