Can I capture an object as a const when using init caputre?

Question

When using C++14's init capture, 'auto' type deduction rules are applied (which throws away the reference, const and volatile)

Notice here that I'm not saying const reference. If I wanted a const reference I can write:

auto lambda = [&widget = std::as_const(widget)] () {};
// C++17

Or I can add & to lhs cast the rhs as a const reference.

what I'm asking for is: can I write something like this?:

auto lambda = [const widget = widget] () {};

Answer 1

There is no way to capture an object by value and have it be const. The grammar for an init-capture is

init-capture:
    identifier initializer
    & identifier initializer

where identifier is

identifier:
    identifier-nondigit
    identifier identifier-nondigit
    identifier digit

Which means you can only specify a name, without any cv-qualifers. For a non mutable lambda this isn't an issue since the function will be const and you can't modify a capture by value object.

For a mutable lambda you can capture a reference to const like you do with [&widget = std::as_const(widget)]. If you can't or don't want to have a reference then you need a write a const wrapper like

template <typename T>
class const_wrapper
{
    const T obj;
public:
    const_wrapper(T obj) : obj(obj) {}
    operator const T&() const { return obj; }
};

The above object is copyable, but not assignable (because of the const member) and only allows const access to the underlying type.