Stream object directly into a std::string

Question

Given some type that is streamable:

struct X {
    int i;

    friend std::ostream& operator<<(std::ostream& os, X const& x) {
        return os << "X(" << x.i << ')';
    }
};

I want to append this onto a std::string. I can implement this as:

void append(std::string& s, X const& x) {
    std::ostringstream os;
    os << x;
    s.append(os.str());
}

But this seems lame since I'm writing data into one stream just to then allocate a new string just for the purposes of appending it onto a different one. Is there a more direct route?

Answer 1

This can be solved by a new type of streambuf (see Standard C++ IOStreams and Locales: Advanced Programmer's Guide and Reference).

Here is a sketch of how it can look:

#include <streambuf>

class existing_string_buf : public std::streambuf
{
public:
    // Store a pointer to to_append.
    explicit existing_string_buf(std::string &to_append); 

    virtual int_type overflow (int_type c) {
        // Push here to the string to_append.
    }
};

Once you flesh out the details here, you could use it as follows:

#include <iostream>

std::string s;
// Create a streambuf of the string s
existing_string_buf b(s);
// Create an ostream with the streambuf
std::ostream o(&b);

Now you just write to o, and the result should appear as appended to s.

// This will append to s
o << 22;

Edit

As @rustyx correctly notes, overriding xsputn is required for improving performance.

Full Example

The following prints 22:

#include <streambuf>
#include <string>
#include <ostream> 
#include <iostream>

class existing_string_buf : public std::streambuf
{
public:
    // Somehow store a pointer to to_append.
    explicit existing_string_buf(std::string &to_append) : 
        m_to_append(&to_append){}

    virtual int_type overflow (int_type c) {
        if (c != EOF) {
            m_to_append->push_back(c);
        }
        return c;
    }

    virtual std::streamsize xsputn (const char* s, std::streamsize n) {
        m_to_append->insert(m_to_append->end(), s, s + n);                                                                                 
        return n;
    }

private:
    std::string *m_to_append;
};


int main()
{   
    std::string s;
    existing_string_buf b(s);
    std::ostream o(&b);

    o << 22; 

    std::cout << s << std::endl;
}   

Answer 2

You could write a std::string casting operator:

struct X {
int i;

friend std::ostream& operator<<(std::ostream& os, X const& x) {
    os << "X(" << x.i << ')';
    return os;
}

operator std::string() {
    return std::string("X(") + std::to_string(x.i) + ")";
}

};

Then, you could simply append it to a std::string:

X myX;
myX.i = 2;
std::string s("The value is ");
s.append(myX); //myX is cast into the string "X(2)"

Answer 3

In this specific case I'd just follow the KISS principle:

struct X {
    int i;

    std::string toString() const {
        return "X(" + std::to_string(i) + ")";
    }
};

Usage:

string += x.toString();
std::cout << x.toString();

An operator<<(std::ostream&, …) isn't really suitable for generic string conversion, so if that's what you're after then a toString type of method / free function is much better. And if you want std::cout << x you can trivially implement operator<< to just call toString.