How does the “lazy evaluation” of Boost Log's

Question

[Follows up Check boost::log filter explicitly? ]

The following example uses the trivial logger from Boost Log. It outputs 1, showing that expensive() is only called once. How does it work? Why is expensive() not called?

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#include <iostream>

#include <boost/log/expressions.hpp>
#include <boost/log/trivial.hpp>

int count = 0;

int expensive()
{
    return ++count;
}

int main()
{
    boost::log::core::get()->set_filter(
        boost::log::trivial::severity >= boost::log::trivial::warning
    );

    BOOST_LOG_TRIVIAL(error) << expensive();
    BOOST_LOG_TRIVIAL(info) << expensive();

    std::cout << count << '\n';

    return 0;
}

Output:

[2018-05-21 14:33:47.327507] [0x00007eff37aa1740] [error]   1
1

Answer 1

It works by using a bit of macro/preprocessor magic. The statements look indeed like a function call to some operator<<():

BOOST_LOG_TRIVIAL(warning) << expensive();

However, simplifying a lot, the macro works as if we had written something like:

if (level == warning)
    logger << expensive();

If you wanted to simplify that code to avoid writing all the time, you could define a macro like this:

#define LOG_WARNING if (level == warning) logger

And then we could use it as:

LOG_WARNING << expensive();

---

The actual BOOST_LOG_TRIVIAL macro ends up expanding into:

for (
    ::boost::log::record _boost_log_record_N =
        (::boost::log::trivial::logger::get()).open_record(
            (::boost::log::keywords::severity = ::boost::log::trivial::error)
        )
    ;
    !!_boost_log_record_N;
)
    ::boost::log::aux::make_record_pump(
        (::boost::log::trivial::logger::get()),
        _boost_log_record_N
    ).stream() << expensive();

As you see, depending on the !!_boost_log_record_N loop condition (which in turn depends on the result of open_record()), the body of the loop will be run zero or more times; which is the reason why expensive() is not always run.