Implicit Conversion Operator for Class Objects to Strings

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I'm learning more about implicit conversion operators, and I've noticed something strange with implicit user-defined conversions for strings. Found below is the code.

#include <iostream>
#include <string>

using std::cout;
using std::endl;
using std::string;

class A {
    public:
        A() {
            *value = "YIKERS";
        } 
        operator string&() {
            return *this->value;
        }
        string* value;
};


int main() {
    A abc;
    cout << abc << endl; 
// compile error: invalid operands to binary expression ('std::__1::ostream' (aka
// 'basic_ostream<char>') and 'A')
    return 0;
}

Any ideas as to why I'm getting this compilation error? I'm thinking this could mean that the object isn't being implicitly converted to a string? If so, why not? Is there a way to fix this? The conversion operator works perfectly fine for other data types like int, float, char, etc.

conversion-operator implicit-conversion operator-overloading c++
1 Answers

First of all, I think you're not using std::string value as intended. This causes another compile error (at least on gcc 10.2). It seems like you want a string and you're using a pointer to a string. This can be fixed by replace string* value with string value, operator string&() with operator string() and *value = "YIKERS' with value = "YIKERS". For the last one you might also want to check initializer lists.

Regarding the current compile error:

The compile error is caused by the code cout << abc trying to use operator<< on an abc which is an object of type A. However, you did not overload this operator. In your example this could be something like

friend std::ostream &operator<<(std::ostream &output, const A &a ) 
{ 
    output << a.value;
    return output;            
}

Even if you have a user-defined conversion to std::string you would still get compile time errors. This link explains it better than I think I could Why cannot use cout with user-defined conversion to std::string?

This is how I understand the explanation from link above:

string header defines the following overload for operator<< for std::basic_ostream:

template <class CharT, class Traits, class Allocator>
std::basic_ostream<CharT, Traits>&
    operator<<(std::basic_ostream<CharT, Traits>& os,
               const std::basic_string<CharT, Traits, Allocator>& str);

Also, std::string is actually a typedef for std::basic_string<char>.

However, if the conversion is implicit the following rule kicks in:

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

So, the compiler cannot deduce the second parameter for the function (e.g. const std::basic_string<CharT, Traits, Allocator>& str) from A. It can, however, deduce it for string so you can explicitly convert abc to string like static_cast<string>(abc) and that would work.

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