scoping issues (was help with eval)

You intended quote() rather than expression(), I believe.  If I do
show.a <- function(a) {a}
eval(quote(show.a(a)),envir=my.env)
this works.

R has lexical scoping, so with
show.a <- function() {a}
'a' is looked for in the frame of the function (not defined there) and 
then in the environment of the function.
environment(show.a)
<environment: R_GlobalEnv>

since show.a was defined in the workspace.  The envir arg of eval() is 
used to find the object(s), here `show.a', not to sets its environment.
So my first version works because 'a' is part of the expression.

Perhaps you intended something like

init.env <- function() {
  a <- 200
  show.a <- function() {a}
  environment()
}
my.env <- init.env()
eval(quote(show.a()),envir=my.env)

which works, and is a common R idiom (although via local() or new.env()).

However, you could just do
eval(quote(a), envir=my.env)
[1] 200
eval(expression(a), envir=my.env)
[1] 200

BTW, use typeof() to see the difference here.

I've been looking at the help page for eval for a while, but I can't
make sense of why this example does not work.

show.a <- function() {
 a
}

init.env <- function() {
 a <- 200
 environment()
}

my.env <- init.env()

ls(envir=my.env)

# returns this:
# > ls(envir=my.env)
# [1] "a"

# but this does not work:
eval(expression(show.a()),envir=my.env)

# > eval(expression(show.a()),envir=my.env)
# Error in show.a() : Object "a" not found
# >

The help page gives the following:

  'eval' evaluates the expression 'expr' argument in the environment
    specified by 'envir' and returns the computed value. If 'envir' is
    not specified, then 'sys.frame(sys.parent())', the environment
    where the call to 'eval' was made is used.

I would be grateful for any help.

Thanks,
Whit

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Brian D. Ripley,                  ripley at stats.ox.ac.uk
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scoping issues (was help with eval)

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