static_assert that checks if template parameter type is accepted by another function

Question

I have a function template defined as follows:

template <typename T>
Test &operator<<(const T &data)
{
    std::cout << data << std::endl;
    return *this;
}

As you can see, I print out to the console data using std::cout, which type is std::ostream. However, T may not be a type handled by std::ostream::operator<<.

That's why I'd like to add a static_assert to check whether T is accepted by the insertion operator or not. Currently, I must do a check for every type that std::ostream::operator<< accept:

template <typename T>
Test &operator<<(const T &data)
{
    static_assert(std::is_arithmetic<T>::value ||
                  std::is_same<T, std::streambuf*>::value ||
                  ...
                  std::is_same<T, std::string>::value, "data type must be compliant to std::ostream::operator<< accepted types");

    std::cout << data << std::endl;
    return *this;
}

Is there a type trait that automatically do that? What would you do?

First off, you can't print a type. So, `std::cout << T << std::endl` is wrong in the first place. What you probably meant is `std::cout << data << std::endl;`. Second I don't see the point in this. If `T` isn't printable `std::cout` will complain and you'll get the error you want. — 101010, Aug 08 '14 at 12:16
Yes sorry I meant `std::cout << data << std::endl` of course, my mind was probably elsewhere when I wrote it. — GuiTeK, Aug 08 '14 at 12:26
@GuiTeK Do you have any particular need to catch that error? If `operator<<` is not overloaded for `T`, the compilation will fail anyway (due to `static_assert` or overload resolution). — edmz, Aug 08 '14 at 12:36

Jonathan Wakely · Accepted Answer · 2014-08-08T13:36:45.783

2

Why do you want a static_assert? Is it acceptable to simply make the function not callable if the insertion isn't valid?

template <typename T>
auto operator<<(const T &data) -> decltype(std::cout << data, std::declval<Test&>())
{
    std::cout << data << std::endl;
    ...
}

If you really want a trait see is_stream_insertable

edited Aug 08 '14 at 13:36

answered Aug 08 '14 at 12:35

Jonathan Wakely

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Undefined behavior here, for reaching the closing brace of a non-void function. – Ben Voigt Aug 08 '14 at 13:34
Yes, thanks for the pedantry, it was copied from the OP. Did you fail to notice it there? Twice? – Jonathan Wakely Aug 08 '14 at 13:35
You've also used the comma operator with a type, which is brokenness not in the original function. – Ben Voigt Aug 08 '14 at 13:36
I think you're capable of writing code to a higher standard than most people asking questions. – Ben Voigt Aug 08 '14 at 13:36
I don't know how to return `Test&` because that type isn't even in the OP's question, so I kept it as close as possible to the original (broken) example – Jonathan Wakely Aug 08 '14 at 13:37
Almost surely `return *this;` – Ben Voigt Aug 08 '14 at 13:37
Indeed it's `return *this;`. I forgot it, sorry. Added. – GuiTeK Aug 08 '14 at 17:55

static_assert that checks if template parameter type is accepted by another function

1 Answers1