Consider the following program:

    #include <iostream>
    #include <ostream>
    #include <vector>
    #include <valarray>
    #include <algorithm>
    #include <iterator>
    template<typename Array>
    void print(const Array& a)
    {
    using namespace std;
    typedef typename Array::value_type T;
    copy(&a[0], &a[0] + a.size(),
    ostream_iterator<T>(std::cout, " "));
    }
    template<typename T, unsigned N>
    unsigned size(T(&)[N]) { return N; }
    int main()
    {
    double array[] = { 0.89, 9.3, 7, 6.23 };
    std::vector<double> v(array, array + size(array));
    std::valarray<double> w(array, size(array));
    print(v); // #1
    std::cout << std::endl;
    print(w); // #2
    std::cout << std::endl;
    }

While the call numbered #1 succeeds, the call numbered #2 fails because the const version of the member function valarray<T>::operator[](size_t) returns a value instead of a const-reference. That seems to be so for no apparent reason, no benefit. Not only does that defeats users' expectation but it also does hinder existing software (written either in C or Fortran) integration within programs written in C++. There is no reason why subscripting an expression of type valarray<T> that is const-qualified should not return a const T&.