Jiří Lechner Jiří Lechner - 2 months ago 33
C++ Question

trivial allocator aware container?

I am studying/playing with allocators trying to understand how it works. But I run into problems trying to implement trivial container that accepts an allocator. For now I ended up with this:

template<class T, class Allocator =std::allocator<T>> class Container {
public:
using allocator_type = Allocator;
using value_type = T;
using pointer = typename std::allocator_traits<allocator_type>::pointer;
using reference = value_type&;
using size_type = std::size_t;

Container( size_type n =0 , const allocator_type& allocator =allocator_type() ){
std::cout << "ctor" << std::endl;
allocator.allocate(n);
};
};

int main(int argc, const char* argv[]){
Container<int> c {5};
return 0;
}


It gives me an error
member function 'allocate' not viable: 'this' argument has type 'const allocator_type' (aka 'const std::__1::allocator<int>'), but function is not marked const


How to fix that error, please? Am I missing something ?
I intend to use traits later but would like to make it work using the old way first.

Answer

Your line

allocator.allocate(n);

attempts to call the allocate method of allocator, which is not defined as a const method. If you look, though, the type of allocator is const allocator_type&, that is, a const reference to allocator_type.

How can you use it then? One thing you can usually do with a const object (or reference to one) is to construct a different non-const object from it. This, for example, builds:

allocator_type(allocator).allocate(n);

As SergeyA correctly notes in the comments, it is fairly common not to construct a temporary ad-hoc allocator_type, but rather make such a member:

    allocator_type m_alloc; // Should probably be private

    Container( size_type n =0 , const allocator_type& allocator =allocator_type() ) : 
            m_alloc{allocator}{
        std::cout << "ctor" << std::endl;
        m_alloc.allocate(n);
    };
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