Does my overload of operator<< (stream insertion) with a left operand not being an ostream is valid

Question

Consider the context, where i want to write to another file descriptor than stdout. Here, My "write" method prints to stdout, but only for understanding.

THIS IS THE MAIN PROBLEM I AM TRYING TO SOLVE:

I have a write(char*) method, i want to convert to c++ style stream through operator<<, to be able to write in C++, obfuscating multiple explicit calls of write, allowing to write in a single line of code without having to concatenate or any other solution.

PERIOD.

Though, on internet, there is nowhere to find an example that overloads operator<< with anything else than ostream on left operand. does that mean, my following example (which compiles! i tested it on a codepen-like website) is a rare case of working overload without ostream, or did i misconcepted the whole thing and "Ter" should be an ostream by inheritance ?

EDIT: How can this be a duplicate ? I am legitimately intrigued why both the topic you linked, and https://arne-mertz.de/2015/01/operator-overloading-common-practice/ or many others never talk about a left operand from another type than ostream.

EDIT2: I am having trouble to be precise on my question. Is the following code a good practice, if not, what should i do to solve the problem i braced with capital letters. Thank you for your answers

#include <iostream>

struct Ter {
    void write(const std::string &str) { std::cout << str; }; //Here we should do something else in our context
    friend Ter &operator<<(Ter &a, std::string &b);
};


Ter &operator<<(Ter &a, const std::string &b)
{
   a.write(b);
   return (a);
}


int main()
{
    Ter toto;
    toto << "b" << "c";
}

It depends on what you want `operator<<` to mean for `Ter`. In this case, it sure looks like it wants to be an `std::ostream`. — François Andrieux, Oct 10 '18 at 17:55
Typically, when you overload an operator, you should do what it's built in counterpart does. In C++ `>>` and `<<` is either bitwise operation or streaming. If you want to concatenate something then `+` and `+=` are used. — NathanOliver, Oct 10 '18 at 17:56
Thank you NathanOlivier, this is right. I understand for the concatenation operators, however, concatenation is a WAY of solving the problem, and not the solution we are searching for. We are actually looking to know why nobody replace multiple calls of a method with a stream operator that would call it in a single manner — stalker2106, Oct 10 '18 at 18:01
@NathanOliver Reopened. But well the main principles of operator overloading are at least related. — πάντα ῥεῖ, Oct 10 '18 at 18:03
@πάντα ῥεῖ Indeed they are, but the example of "operator<<" is still written with an ostream as left operand. I dare you to find somewhere on the internet where an exemple of overload of operator<< exists with a left operand of T type that is not ostream& — stalker2106, Oct 10 '18 at 18:04
@stalker2106 Concatenation might be better expressed using an overload of `operator+()`, though `std::string` concatenation might be easily achieved as a combination of overloading `operator<<()` and using `std::ostringstream`. — πάντα ῥεῖ, Oct 10 '18 at 18:06
@πάντα ῥεῖ I don't want to concatenate, but call write two times internally with the strings i supplied in the stream operation. Am I right ? — stalker2106, Oct 10 '18 at 18:09
@stalker2106: "*I dare you to find somewhere on the internet where an exemple of overload of operator<< exists with a left operand of T type that is not ostream&*" [Boost.Spirit](http://boost-spirit.com/home/). Also [Boost.Serialization](https://www.boost.org/doc/libs/1_67_0/libs/serialization/doc/index.html); archives [are not streams](https://www.boost.org/doc/libs/1_67_0/libs/serialization/doc/rationale.html#archives). Also Boost's dynamic bitset type. — Nicol Bolas, Oct 10 '18 at 18:10
@Nicol Bolas On every link you supplied, when i am looking for "<<" in my browser search, nothing shows! Where can I find an example of an overload on these pages ? — stalker2106, Oct 10 '18 at 18:15
It's difficult to read this as a clear, specific question. _"Why do people always..."_ They don't always. _"There is nowhere to find..."_ There is. _"Does that mean my example is a rare case..."_ No, it doesn't mean that. — Drew Dormann, Oct 10 '18 at 18:16
@stalker2106: I didn't link you to code; I linked you to *libraries*. Feel free to find the code within those libraries yourself. — Nicol Bolas, Oct 10 '18 at 18:16
Can you clarify what kind of answer you're looking for? The new question in your title is answered in the question itself. _You_ provide an example. Are you looking for links to popular libraries? Or are you asking if your code is wrong? — Drew Dormann, Oct 10 '18 at 18:23
@Drew Dormann I am actually CTRL+F-ing Boost sources as suggested by Nicol Bolas an I managed to find examples which is nice and partially answers my question telling me this is 'valid'. I also edited the question to be more precise. — stalker2106, Oct 10 '18 at 18:31
your question is rather vague. If you have working code and you are not sure if it is a good way to write it you may have more luck on https://codereview.stackexchange.com — 463035818_is_not_an_ai, Oct 10 '18 at 18:39

score 1 · Accepted Answer · answered Oct 10 '18 at 18:43

1

I am legitimately intrigued why both the topic you linked, and https://arne-mertz.de/2015/01/operator-overloading-common-practice/ or many others never talk about a left operand from another type than ostream.

Because it does not really matter. What works for std::ostream works similar for other types. Sorry, but its a bit like getting an example of overloading operator+ for a type Foo and then complaining that you never get an example of how to overload it for a Bar. If you want your type to offer a operator<< then yes, you implemented it in way that works.

Your code seems to be correct. Whether it does what you want, only you can decide.

answered Oct 10 '18 at 18:43

463035818_is_not_an_ai

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This is a good answer thank you. Yeah well, I'm just intrigued why (for example in the link i supplied) all the operators are overloaded for a T or X type, and operator<< is always with an ostream. and i find your answer very relevant because you show a use case, and with + operator anybody understands x + y, but people wont for a << b. thank you. – stalker2106 Oct 10 '18 at 18:48
@stalker Thats because typically you do care about the right hand side and do want to supply an overload for `std::ostream`s `operator<<` – 463035818_is_not_an_ai Oct 10 '18 at 19:04

Does my overload of operator<< (stream insertion) with a left operand not being an ostream is valid

1 Answers1