On Tue, Mar 01, 2005 at 11:21:44PM -0800, Seth Falcon wrote:
On Feb 25, 2005, at 12:34 PM, [EMAIL PROTECTED] 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)
On Mar 2, 2005, at 4:13 AM, Jan T. Kim wrote:
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
On Feb 25, 2005, at 12:34 PM, [EMAIL PROTECTED] 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
Is is possible from within a function to cause its caller to return()?
I have a function that lets user make edits to certain objects, and then
checks that the edited objects still make sense. If they don't, the function
puts up a notifier that the edits are being discarded and then returns,
On Fri, 25 Feb 2005 [EMAIL PROTECTED] wrote:
Is is possible from within a function to cause its caller to return()?
Not as such. You probably want to signal and catch a condition. Look at
?tryCatch.
-thomas
I have a function that lets user make edits to certain objects, and then
checks
