The note refers to the fact that the function named ‘fun’ appears to be defined
in two different ways.
From: Hervé Pagès <hpages.on.git...@gmail.com>
Sent: Tuesday, February 6, 2024 2:17 PM
To: Duncan Murdoch <murdoch.dun...@gmail.com>; Izmirlian, Grant (NIH/NCI) [E]
<izmir...@mail.nih.gov>; r-devel@r-project.org
Subject: [EXTERNAL] Re: [Rd] NOTE: multiple local function definitions for
?fun? with different formal arguments
Thanks. Workarounds are interesting but... what's the point of the NOTE in the
first place?
H.
On 2/4/24 09:07, Duncan Murdoch wrote:
On 04/02/2024 10:55 a.m., Izmirlian, Grant (NIH/NCI) [E] via R-devel wrote:
Well you can see that yeast is exactly weekday you have. The way out is to
just not name the result
I think something happened to your explanation...
toto <- function(mode)
{
ifelse(mode == 1,
function(a,b) a*b,
function(u, v, w) (u + v) / w)
}
It's a bad idea to use ifelse() when you really want if() ... else ... . In
this case it works, but it doesn't always. So the workaround should be
toto <- function(mode)
{
if(mode == 1)
function(a,b) a*b
else
function(u, v, w) (u + v) / w
}
________________________________
From: Grant Izmirlian <izmirlidr...@gmail.com><mailto:izmirlidr...@gmail.com>
Date: Sun, Feb 4, 2024, 10:44 AM
To: "Izmirlian, Grant (NIH/NCI) [E]"
<izmir...@mail.nih.gov><mailto:izmir...@mail.nih.gov>
Subject: Fwd: [EXTERNAL] R-devel Digest, Vol 252, Issue 2
Hi,
I just ran into this 'R CMD check' NOTE for the first time:
* checking R code for possible problems ... NOTE
toto: multiple local function definitions for �fun� with different
formal arguments
The "offending" code is something like this (simplified from the real code):
toto <- function(mode)
{
if (mode == 1)
fun <- function(a, b) a*b
else
fun <- function(u, v, w) (u + v) / w
fun
}
Is that NOTE really intended? Hard to see why this code would be
considered "wrong".
I know it's just a NOTE but still...
I agree it's a false positive, but the issue is that you have a function object
in your function which can't be called unconditionally. The workaround doesn't
create such an object.
Recognizing that your function never tries to call fun requires global
inspection of toto(), and most of the checks are based on local inspection.
Duncan Murdoch
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Hervé Pagès
Bioconductor Core Team
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