Unexpected behaviour with argument defaults

Question

I just ran into something odd, which hopefully someone here can shed some light on. Basically, when a function has an argument whose default value is the argument\'s name, s

Answer 1

Default arguments are evaluated inside the scope of the function. Your f2 is similar (almost equivalent) to the following code:

f2 = function(y) {
    if (missing(y)) y = y
    y^2
}

This makes the scoping clearer and explains why your code doesn’t work.

Note that this is only true for default arguments; arguments that are explicitly passed are (of course) evaluated in the scope of the caller.

Lazy evaluation, on the other hand, has nothing to do with this: all arguments are lazily evaluated, but calling f2(y) works without complaint. To show that lazy evaluation always happens, consider this:

f3 = function (x) {
    message("x has not been evaluated yet")
    x
}

f3(message("NOW x has been evaluated")

This will print, in this order:

x has not been evaluated yet
NOW x has been evaluated