Why doesn't a conversion function work with std::string?

Question

Consider the following class which contains a conversion function for the std::string type:

class SomeType
{
    public:

    SomeType(char *value)
    {
        _str = value;
    }

    operator std::string()
    {
        return std::string(_str);
    }

    private:
    char *_str;
};

The following snippet fails to compile with the error: no operator "==" matches these operands

int main(int argc, char* argv[])
{
    SomeType a("test");

    if (a == std::string("test")) // ERROR on this line
    {
        int debug = 1;
    }

    return 0;
}

I realize I could define an operator== method that accepts std::string operand, but why doesn't the conversion function work?

@Aniket The error is in the question. – Konrad Rudolph Jul 15 '13 at 00:05 — Konrad Rudolph, Jul 15 '13 at 00:05
If any of these answers was helpful, please accept one. – Borgleader Jul 15 '13 at 12:03 — Borgleader, Jul 15 '13 at 12:03

score 8 · Accepted Answer · edited May 23 '17 at 11:56

The problem is that std::string is in fact a template and as such I imagine that it's comparison operator is also a template. And in that case the standard as I recall states that no implicit conversion will happen for the required arguments, which means you'd have to cast SomeType to string yourself for it to be called.

As stated in Item 46 of Effective C++:

[...], because implicit type conversions are never considered during template argument deduction. Never. Such conversions are used during function calls, yes, but before you can call a function, you have to know which functions exist. [...]

You can find more info here.

RiaD · Answer 2 · 2013-07-15T00:38:42.737

4

std::string is actually typedef to std::basic_string<char, i_do_not_remember>

There no operator == that get just std::string It's template one

template<...>
bool operator (basic_string<...>& a, basic_string<...>& b) {
    //
}

But template types can't be deduced here. You may cast it manually:

if (static_cast<std::string>(a) == std::string("test"))

edited Jul 15 '13 at 00:38

answered Jul 15 '13 at 00:07

RiaD

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Why doesn't a conversion function work with std::string?

2 Answers2

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