Comparing function of binary_search - c++

Question

I am trying to run binary_search on vector of custom objects.
struct T{
string name;
T(string n):name(n){};
bool operator < ( T * n ) const {
return name < n -> name;
}
bool operator == ( T * n ) const {
return name == n -> name;
}
};
vector<T *> t;
t.push_back(new T("one"));
t.push_back(new T("two"));
t.push_back(new T("three"));
bool has_3 = binary_search( t.begin(), t.end(), new T("two") ) ;
if( has_3 ){
cout <<"Its there" << endl;
}
The comparation function should be just fine yet when i run the code has_3 equals to 0 = the element isnt present in vector. Is this problem caused by my overloading of < ? I see no reason why this shouldnt find the value. Considering the order of insertion into vector it should be sorted
Thanks for help.

Answer 1

There are several reasons why this shouldn't find the value:
The range must be sorted; your range is out of alphabetical order
Your comparison functionality is defined between T and T*, while you search a vector of T* for a T*.
You can fix the first problem by swapping "two" and "three", and the second problem by making a vector of T:
struct T{
string name;
T(string n):name(n){};
bool operator < ( const T &n ) const {
return name < n.name;
}
// operator == is not necessary for binary_search
};
int main() {
vector<T> t;
t.push_back(T("one"));
t.push_back(T("three"));
t.push_back(T("two"));
bool has_3 = binary_search( t.begin(), t.end(), T("two") ) ;
if( has_3 ){
cout <<"Its there" << endl;
}
return 0;
}
Demo 1.
If you do have no way but to construct a vector of pointers, you have this ugly work-around available (I strongly recommend against it):
struct T{
string name;
T(string n):name(n){};
};
bool operator < (const T& l, const T *r) {
return l.name < r->name;
}
bool operator < (const T *l, const T &r) {
return l->name < r.name;
}
Now you can search like this:
bool has_3 = binary_search( t.begin(), t.end(), T("two") ) ;
if( has_3 ){
cout <<"Its there" << endl;
}
Demo 2.

Answer 2

It's a really dumb requirement to work with a vector of pointers to dynamically allocated objects. But here is an approach that will work.
#include <iostream>
#include <string>
#include <algorithm>
struct T
{
std::string name;
T(std::string n):name(n){};
};
// this is the comparater needed to work with pointers, but it should
// NOT be a member of T
bool pointer_comparer(const T *left, const T *right)
{
// this assumes both left and right point to valid objects
return left->name < right->name;
}
int main()
{
std::vector<T *> t;
t.push_back(new T("one"));
t.push_back(new T("two"));
t.push_back(new T("three"));
// t is unsorted. We need to sort it since binary_search will
// ASSUME it is sorted
std::sort(t.begin(), t.end(), pointer_comparer);
T *value_needed = new T("two");
bool has_3 = std::binary_search( t.begin(), t.end(), value_needed, pointer_comparer);
if(has_3)
{
std::cout <<"Its there" << std::endl;
}
// since we've been stupidly allocating objects, we need to release them
delete value_needed;
for (std::vector<T *>::iterator i = t.begin(), end = t.end();
i != end; ++i)
{
delete (*i);
}
// and since t now contains a set of dangling pointers, we need to discard them too
t.resize(0);
return 0;
}
Why do I say the requirement to work with a vector of pointers to dynamically allocated objects. Compare the above with an approach that works with a vector<T> rather than a vector<T *>.
#include <iostream>
#include <string>
#include <algorithm>
struct T
{
std::string name;
T(std::string n):name(n){};
bool operator < (const T &) const
{
return name < n.name;
};
};
int main()
{
std::vector<T> t;
t.push_back(T("one"));
t.push_back(T("two"));
t.push_back(T("three"));
// t is unsorted. We need to sort it since binary_search will
// ASSUME it is sorted
std::sort(t.begin(), t.end());
bool has_3 = std::binary_search(t.begin(), t.end(), T("two"));
if(has_3)
{
std::cout <<"Its there" << std::endl;
}
// we need do nothing here. All objects use above will be properly released
return 0;
}
Note: I've written the above so it works with ALL C++ standards. Assuming C++11 and later, simplifications are possible in both cases.