How to use the same object instance when creating other objects?

Question

I'm new to C++ and started playing with references, which led me to the following code:

#include <iostream>
#include <unordered_map>

class Wrapper {
private:
    std::unordered_map<std::string, int> map;
public:
    void add(std::string &key, int value) { map[key] = value; }
    int get(std::string &key) { return map[key]; }
};

class Writer {
private:
    Wrapper wrapper;
public:
    explicit Writer(const Wrapper &wrapper) : wrapper(wrapper) 
    { 
        std::cout << "Writer: " << &wrapper << std::endl; 
    }
    void write(std::string key, int value) { wrapper.add(key, value); }
};

class Reader {
private:
    Wrapper wrapper;
public:
    explicit Reader(const Wrapper &wrapper) : wrapper(wrapper) 
    { 
        std::cout << "Reader: " << &wrapper << std::endl; 
    }
    int read(std::string key) { return wrapper.get(key); }
};

My main function:

int main() 
{
    Wrapper wrapper;
    Writer writer(wrapper);
    Reader reader(wrapper);

    writer.write("key", 123);
    std::cout << "Value: " << reader.read("key") << std::endl;
}

I pass a reference to the same instance of the Wrapper class when I create an instance of Writer and Reader. I expected that the value added by the writer should also be available to the reader, since they use the same wrapper instance. However, the read("key") call returns 0 because the key is unknown.

What am I doing wrong and how can I achieve the desired behavior?

`wrapper(wrapper)` is calling the [_copy constructor_](https://stackoverflow.com/questions/2168201/what-is-a-copy-constructor-in-c). The default implementation of a copy constructor will make a _copy_. — Drew Dormann, Aug 08 '22 at 13:58
You pass a reference but `wrapper` is not a reference in your `Reader` and `Writer` classes. Make it so and it should work. — john, Aug 08 '22 at 14:01

score 7 · Accepted Answer · answered Aug 08 '22 at 14:00

7

member variable must also be declared as reference, in both reader and writer

class Writer {
private:
    Wrapper & wrapper;
}


class Reader {
private:
    Wrapper & wrapper;
}

answered Aug 08 '22 at 14:00

Marc Stroebel

2,295
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1

Pedantry, but a pointer would also work. – john Aug 08 '22 at 14:02
1

I don't like mixing pointer and reference syntax... – Marc Stroebel Aug 08 '22 at 14:03
3

Using member variables that are references has serious implications for copy constructor, move constructor, copy assignment, and move assignment. (Meaning that the programmer should be aware of them; not that they shouldn't use reference member variables.) – Eljay Aug 08 '22 at 14:11
1

@Eljay same for a (raw) pointer, but with a pointer the consequences are easier to handle – 463035818_is_not_an_ai Aug 08 '22 at 14:13
1

btw I don't think it is "pedantry". I agree that references as members are rarely the right choice. While this is somewhat opinionated, saying OP "**must** declare the members as reference" isn't right – 463035818_is_not_an_ai Aug 08 '22 at 14:14
sorry, I'm not a native english speaker... perhaps there's a better expression than "must be .." – Marc Stroebel Aug 08 '22 at 17:49
i personally try to use "You can ..." rather than "You must..." – 463035818_is_not_an_ai Aug 08 '22 at 19:59

How to use the same object instance when creating other objects?

1 Answers1