dear Bert,
Thanks a lot...
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
________________________________
From: Bert Gunter <bgunter.4...@gmail.com>
Sent: Monday, January 9, 2023 9:59 PM
To: akshay kulkarni <akshay...@hotmail.com>
Cc: R help Mailing list <r-help@r-project.org>
Subject: Re: [R] return value of {....}
Unless you do something special within a function, only the value(s) returned
are available to the caller. That is the essence of functional-type programming
languages.
You need to read up on (function) environments in R . You can search on this.
?function and its links also contain useful information, but it may too terse
to be explicable to you. There are of course many available references on the
internet.
I believe your mental model for how R works is flawed, and you have some
homework to do to correct it. I may be wrong, naturally, but you can judge by
looking at some tutorials.
-- Bert
On Mon, Jan 9, 2023 at 6:47 AM akshay kulkarni
<akshay...@hotmail.com<mailto:akshay...@hotmail.com>> wrote:
Dear members,
I have the following code:
> TB <- {x <- 3;y <- 5}
> TB
[1] 5
It is consistent with the documentation: For {, the result of the last
expression evaluated. This has the visibility of the last evaluation.
But both x AND y are created, but the "return value" is y. How can this be
advantageous for solving practical problems? Specifically, consider the
following code:
F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10}
Both expr5 and expr7 are created, and are accessible by the code outside of the
nested braces right? But the "return value" of the nested braces is expr7. So
doesn't this mean that only expr7 should be accessible? Please help me entangle
this (of course the return value of F is expr10, and all the other objects
created by the preceding expressions are deleted. But expr5 is not, after the
control passes outside of the nested braces!)
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
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