static_cast on custom class causes copy assignment to fail

Question

I expected the following program to print "11" but it actually prints "01" so it seems like the first assignment fails.

struct A
{
    A(int i = 0) : i_(i) {}
    int i_;
};

int main()
{
    A x(1);
    A y;
    static_cast<A>(y) = x; // *** Fails to assign ***
    std::printf("%i", y.i_);
    y = x;
    std::printf("%i", y.i_);
}

If I use a primitive type likeint instead of A then int x = 1; int y; static_cast<int>(y) = x; does assign the value 1 to x. Is there some way I can get it to work for custom types? I tried adding operator A() { return *this; } to struct A but that didn't work.

Obviously this is a stupid program but the problem arises in a template function where I have static_cast<std::remove_const<T>::type>(y) = x and it was working fine for primitive types but just now failed for a custom type.

`std::remove_const` should be used with *extreme* care and not lightly - the risk of introducing Undefined Behaviour is great. — Jesper Juhl, Sep 19 '16 at 18:49

Lightness Races in Orbit · Accepted Answer · 2016-09-19T18:50:23.637

4

As with any cast, static_cast<A>(y) is a temporary copy of y. You can cast to a reference type instead (static_cast<A&>(y)); more generally, you could achieve this with std::add_lvalue_reference.

For the more specific example you described, you'll need const_cast rather than static_cast, but the basic principle is the same.

Here's an example that compiles, but has UB because of the modification of a const object (and thus returns 0, not 42). Without knowing more about what you're trying to do, I shan't attempt to disguise that for the purposes of this example:

#include <iostream>
#include <type_traits>

template <typename T>
T foo(T val)
{
    T x{};

    using not_const = typename std::remove_const<T>::type;
    using ref_type  = typename std::add_lvalue_reference<not_const>::type;

    const_cast<ref_type>(x) = val;

    return x;
}

int main()
{
    std::cout << foo<const int>(42) << '\n';
}

edited Sep 19 '16 at 18:50

answered Sep 19 '16 at 18:41

Lightness Races in Orbit

378,754
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1,055

Thank you that makes sense. But why does it work for primitive types if the value is only being assigned to a temporary copy? – George Skelton Sep 19 '16 at 18:55
But `int x = 1; int y = 0; static_cast(y) = x; std::cout << y;` prints `1` – George Skelton Sep 19 '16 at 19:00
@GeorgeSkelton: No, it doesn't. It doesn't even compile, at least not in a standard-compliant C++ compiler. Click on the link I just gave you for a live demonstration. – Lightness Races in Orbit Sep 19 '16 at 19:01
Ah ok just saw your code. I'm using VS which doesn't give me a warning and appears to "work" – George Skelton Sep 19 '16 at 19:01

static_cast on custom class causes copy assignment to fail

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