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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T16:27:20+00:00 2026-05-13T16:27:20+00:00

I’m trying to implement my own qDebug() style debug-output stream, this is basically what

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I’m trying to implement my own qDebug() style debug-output stream, this is basically what I have so far:

struct debug
{
#if defined(DEBUG)
    template<typename T>
    std::ostream& operator<<(T const& a) const
    {
        std::cout << a;
        return std::cout;
    }
#else
    template<typename T>
    debug const& operator<<(T const&) const
    {
        return *this;
    }

    /* must handle manipulators (endl) separately:
     * manipulators are functions that take a stream& as argument and return a
     * stream&
     */
    debug const& operator<<(std::ostream& (*manip)(std::ostream&)) const
    {
        // do nothing with the manipulator
        return *this;
    }
#endif
};

Typical usage:

debug() << "stuff" << "more stuff" << std::endl;

But I’d like not to have to add std::endl;

My question is basically, how can I tell when the return type of operator<< isn’t going to be used by another operator<< (and so append endl)?

The only way I can think of to achieve anything like this would be to create a list of things to print with associated with each temporary object created by qDebug(), then to print everything, along with trailing newline (and I could do clever things like inserting spaces) in ~debug(), but obviously this is not ideal since I don’t have a guarantee that the temporary object is going to be destroyed until the end of the scope (or do I?).

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  1. Editorial Team
    Editorial Team
    2026-05-13T16:27:20+00:00Added an answer on May 13, 2026 at 4:27 pm

    Qt uses a method similar to @Evan. See a version of qdebug.h for the implementation details, but they stream everything to an underlying text stream, and then flush the stream and an end-line on destruction of the temporary QDebug object returned by qDebug().

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