Using std :: max_element for a vector <double>

Question

I am trying to use std::min_element and std::max_element to return the minimum and maximum elements in a doubles vector. My compiler does not like how I am currently trying to use them, and I do not understand the error message. I could, of course, write my own procedure to find min / max, but I would like to understand how to use functions.

 #include <vector> #include <algorithm> using namespace std; int main(int argc, char** argv) { double cLower, cUpper; vector<double> C; // code to insert values in C not shown here cLower = min_element(C.begin(), C.end()); cUpper = max_element(C.begin(), C.end()); return 0; } 

Here is the compiler error:

 ../MIXD.cpp:84: error: cannot convert '__gnu_cxx::__normal_iterator<double*, std::vector<double, std::allocator<double> > >' to 'double' in assignment ../MIXD.cpp:85: error: cannot convert '__gnu_cxx::__normal_iterator<double*, std::vector<double, std::allocator<double> > >' to 'double' in assignment 

Can someone please explain what I am doing wrong?

Answer 1

min_element and max_element return iterators, not values. Therefore, you need *min_element... and *max_element...

Answer 2

As others have said, std::max_element() and std::min_element() return the iterators that need to be dereferenced to get the value.

The advantage of returning an iterator (and not just a value) is that it allows you to determine the position of the (first) element in the container with the maximum (or minimum) value.

For example (using C ++ 11 for short):

 #include <vector> #include <algorithm> #include <iostream> int main() { std::vector<double> v {1.0, 2.0, 3.0, 4.0, 5.0, 1.0, 2.0, 3.0, 4.0, 5.0}; auto biggest = std::max_element(std::begin(v), std::end(v)); std::cout << "Max element is " << *biggest << " at position " << std::distance(std::begin(v), biggest) << std::endl; auto smallest = std::min_element(std::begin(v), std::end(v)); std::cout << "min element is " << *smallest << " at position " << std::distance(std::begin(v), smallest) << std::endl; } 

This gives:

 Max element is 5 at position 4 min element is 1 at position 0 

---

Note:

Using std::minmax_element() , as suggested in the comments above, may be faster for large datasets, but may give slightly different results. The values ​​for my example above will be the same, but the position of the "max" element will be 9 , because ...

If multiple elements are equivalent to the largest element, an iterator is returned to the last such element.

Answer 3

min / max_element returns an iterator in the min / max element, not the value of the min / max element. You must dereference the iterator to get the value and assign it to the double. I.e:

 cLower = *min_element(C.begin(), C.end());