c++ left hand assignment operator

Question

i have a class (wich should hold any value) like this:

class Value
{

    public:

        Value();
        Value(const Value& value);
        virtual ~Value();
        void operator= (const Value& value);

        template<class T>
        void operator= (T value);
...
}

now my question:

why can't i implement an assignment operator for this class like this:

template<class T>
void operator=(T& value, const Value& v)
{...}

I wan to desing a class wich works the following:

Value v;

v = 'c';
v = 13;
v = 5.6;
int i = 5;
v = &i;

int y = v;
char b = v;

i want to put any datatype into it and out of it. at the moment this works fine for:

v = 'c';
v = 13;
v = 5.6;

but not for:

int y = v;

what works is:

int y = v.get<int>();

but this is not as nice as

int y = v;

would be

Is that a typo in your question? Did you mean `void operator=`? — melpomene, Oct 02 '15 at 14:13
See [Overloading assignment operator in a class template that can cast to another template type](http://stackoverflow.com/questions/8305952/overloading-assignment-operator-in-a-class-template-that-can-cast-to-another-tem). — agold, Oct 02 '15 at 14:15
Why would you want to implement it like this? What problem do you believe it would help you solve? — Igor Tandetnik, Oct 02 '15 at 14:24
Just call the function `assign` instead of `operator=` and all will be good. If not, then you have a much more serious design problem anyway. — Christian Hackl, Oct 02 '15 at 14:27
BTW , it's expected that `operator=` will return a reference to `*this`, so its signature should be `Value& operator=(T)`. — Toby Speight, Oct 02 '15 at 14:30
Look into Boost::Any or Boost::Variant unless you're just doing this for fun. — AndyG, Oct 02 '15 at 14:52

score 0 · Answer 1 · answered Oct 02 '15 at 14:22

0

Because the standard says that an assignment operator must be a member function with only one parameter.

13.5.3$1 Assignment [over.ass]:

An assignment operator shall be implemented by a non-static member function with exactly one parameter.

answered Oct 02 '15 at 14:22

songyuanyao

169,198
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score 0 · Answer 2 · answered Oct 02 '15 at 14:49

0

You can implement a typecast operator like this

operator int()
{
    if(current_value_is_not_int)
        throw MyException("Current value is not int");

    //return int value
}

answered Oct 02 '15 at 14:49

Thomas Sparber

2,827
2
18
34

score 0 · Accepted Answer · answered Oct 02 '15 at 14:54

You can easily fix the compilation error by adding template type cast to your class like following:

class Value
{
...
        template <class T> operator T();
};

Value va;
int i = va;

I still believe you will find the task of implementing 'boost::any' yourself quite challenging, but why not? :)

c++ left hand assignment operator

3 Answers3