Why does evaluating an expression in system.time() make variables available in global environment?

Question

Can somebody please explain what happens when an expression is evaluated in system.time? In particular, why are any variables that are declared in the ex

Answer 1

EDIT : as per @Tommy's comment: The evaluation actually only takes place once the argument expr is used (that's the lazy evaluation).

What is passed is a language object, not an expression. You basically nest the <- function (with two arguments) within the st() function call, and the result of the <- call is passed to to st. As you can see in ?assignOps, the <- function returns the assigned value silently. As @Josh told you already, this evaluation of the nested function takes place in the environment the function is called from.

What you do, is equivalent to

st(mean(1:10))

To see the difference, you can do:

st <- function(expr){
  typeof(expr)
}
> st(aa <- 1)
[1] "double"
> st(expression(aa <- 1))
[1] "expression"

For the structure of the call, you can do:

st <- function(expr){
  str(as.list(match.call()))
}
> st(mean(1:10))
List of 2
 $     : symbol st
 $ expr: language mean(1:10)
> st(aa <- 1)
List of 2
 $     : symbol st
 $ expr: language aa <- 1