有没有更安全的方法将整数连接到 const char？

Question

consider a class called foo

class foo
{
    int i;
    operator const char*()
    {
        char buf[255];
        sprintf(buf, "your number is: %i", i);
        return buf;
    }
};

and I want to cast it to const char*. What is the best way to do it? Right now my solution shows it by using sprintf. But I am afraid that when the function returns, the section of the code will be marked as freed, and will be overwritten...

I have also tried by using std::string, in which I just return var.c_str(). But by the time function terminates var destructs and I just get random output all over the place...

Am I just being over-exaggerating right now or is there a better way of doing it?

Note1: I know that by returning a variable of type std::string containing the stuff is a viable solution but it won't be immediately recognized by cout or printf (requires casting the foo variable).

if I were to use a std::string conversion,

class foo
{
private:
    int i;
public:
    foo() : i(9) {};
    operator std::string()
    {
        return std::string(std::to_string(i));
    }
};

int main()
{
    foo bar = foo();
    std::cout << bar; // no operator "<<" matches these operands, operand types are: std::ostream << foo
}

Answer 1

Your code has undefined behavior as it returns a pointer to a local variable that goes out of scope when the operator exits. You need to make the char[] buffer outlive the operator.

You can make the buffer be a member of the class:

class foo {
    int i;
    char buf[255];
    operator const char*() const {
        sprintf(buf, "your number is: %i", i);
        return buf;
    }
};

Or, you can make the buffer be static :

class foo {
    int i;
    operator const char*() const {
        static char buf[255];
        sprintf(buf, "your number is: %i", i);
        return buf;
    }
};

A better option is to return a std::string instead:

class foo {
    int i;
    operator std::string() const {
        return "your number is: " + std::to_string(i);
    }
};

But then you have to cast the bar variable, as you already discovered:

std::cout << static_cast<std::string>(bar);

Or assign it to a variable first:

std::string s = bar;
std::cout << s;

A cleaner solution is to define an overload of operator<< for foo :

class foo {
    int i;
    void print(std::ostream &out) const {
        out << "your number is: " << i;
    }
    operator std::string() const {
        std::ostringstream oss;
        print(oss);
        return oss.str();
    }
};

std::ostream& operator<<(std::ostream &out, const foo &f) {
    f.print(out);
    return out;
}

std::cout << bar; // now it works!