Using std::max_element on a vector<double>

Question

I'm trying to use std::min_element and std::max_element to return the minimum and maximum elements in a vector of doubles. My compiler doesn't like how I'm currently trying to use them, and I don't understand the error message. I could of course write my own procedure to find the minimum and maximum, but I'd like to understand how to use the 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 is 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 

What am I doing wrong?

Answer 1

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

Answer 2

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

The advantage of returning an iterator (rather than just the 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 brevity):

#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 yields:

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

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Note:

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

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