return from nested function?

On Feb 25, 2005, at 12:34 PM, jhallman at frb.gov wrote:

Is is possible from within a function to cause its caller to return()?

This snippet may be of interest:

f = function(x) {

+ print("f")
+ g(return())
+ print("end of f")
+ }

g=function(x) {print("g")

+ x
+ print("end of g")
+ }

f(1)

[1] "f"
[1] "g"
NULL

I may be dumb today, but doesn't that beg the question of how does g
cause f not to return?

Best regards, Jan

+- Jan T. Kim -------------------------------------------------------+
 |    *NEW*    email: jtk at cmp.uea.ac.uk                               |
 |    *NEW*    WWW:   http://www.cmp.uea.ac.uk/people/jtk             |
 *-----=<  hierarchical systems are for files, not for humans  >=-----*

On Wed, 2 Mar 2005 12:13:17 +0000, "Jan T. Kim" <jtk at cmp.uea.ac.uk>
wrote :

On Tue, Mar 01, 2005 at 11:21:44PM -0800, Seth Falcon wrote:

On Feb 25, 2005, at 12:34 PM, jhallman at frb.gov wrote:

Is is possible from within a function to cause its caller to return()?

This snippet may be of interest:

f = function(x) {

+ print("f")
+ g(return())
+ print("end of f")
+ }

g=function(x) {print("g")

+ x
+ print("end of g")
+ }

f(1)

[1] "f"
[1] "g"
NULL

I may be dumb today, but doesn't that beg the question of how does g
cause f not to return?

I'm not suggesting that I would ever use code like this, but if x is
never evaluated then the double return would not happen:

f

function(x) {
print("f")
g(x, return())
print("end of f")
}

g

function(x, blastoff) {
   print("g")
   if (x) blastoff
   print("end of g")
}

f(TRUE)

[1] "f"
[1] "g"
NULL

f(FALSE)

[1] "f"
[1] "g"
[1] "end of g"
[1] "end of f"

However, this is not a style of programming that I would expect to
last for more than 5 minutes after I published a program using it.
Follow Thomas's advice, and use tryCatch instead.

Duncan Murdoch

I may be dumb today, but doesn't that beg the question of how does g
cause f not to return?

No, I think my post doesn't really help you.  In the context of a 
recursive function, the code I posted provides a way to jump out of the 
recursion which is a cousin of what you are looking for.

I wonder if try or tryCatch might be what you want?

cheers,

+ seth