The message will not be displayed.
The ReferenceError error will appear, because here not a simple functional expression is used, but a named functional expression in which the function name is visible only inside the function itself.
A variable with this name is not created outside the function.
By the way, you can still get this name outside the function: you need to use name
property.
In this case, console.log(f.name);
will write 'coolFunction'
.