writeln() and format() with variadic templates
The biggest thing that sticks out at me is that you're passing all of your Args... parameters by value, thus pretty much making std::forward useless here. To make use of std::forward, the reference type needs to be deduced from the calling context. By itself, std::forward really doesn't do anything except a static_cast to the deduced reference type.
Basically, everywhere you have Args should be passed by Args&amp;&amp;:
template&lt;typename T, typename ...Args&gt;
void writeln(T firstArg, Args&amp;&amp;... extraArgs)

template&lt;typename T, typename ...Args&gt;
std::string format(T firstArg, Args&amp;&amp;... extraArgs)

As a cut down example of the difference this makes, try this example:
#include &lt;memory&gt;
#include &lt;iostream&gt;

struct s
{
    s()
    { }
    
    s(s&amp;&amp; )
    {
        std::cout &lt;&lt; &quot;Move\n&quot;;
    }
    
    s(const s&amp; )
    {
        std::cout &lt;&lt; &quot;Copy\n&quot;;
    }
};

template &lt;typename T&gt;
void do_forward(T&amp;&amp; v)
{
    sink(std::forward&lt;T&gt;(v));
}

template &lt;typename U&gt;
void sink(U&amp;&amp; u)
{
    std::cout &lt;&lt; &quot;In sink\n&quot;;
}

int main()
{
    s something;
    do_forward(something);
}

If I run this, the output is:

In sink

If I change the signature of do_foward and sink to:
template &lt;typename T&gt;
void do_forward(T v)

template &lt;typename U&gt;
void sink(U u)

The output is:

Copy
Move
In sink

If we change main to just pass an actual rvalue reference instead of an lvalue:
int main()
{
    do_forward(s{});
}

We remove the first copy operation (from do_forward) but still have a move operation (in sink).