[R] Default argument not passed to subfunction when argument name matches default expression

Hello, I have difficulties to understand this one: foo - function (y = 2) { bar - function (y = y) y^2 bar() } foo() #! Error in y^2 : 'y' is missing foo(3) #! Error in y^2 : 'y' is missing Note that this one works: foo - function (y = 2) { bar - function (y = y) y^2 bar(y) #

Re: [R] Default argument not passed to subfunction when argument name matches default expression

On 31/03/2014 10:40 AM, Philippe Grosjean wrote: Hello, I have difficulties to understand this one: foo - function (y = 2) { bar - function (y = y) y^2 bar() } foo() #! Error in y^2 : 'y' is missing foo(3) #! Error in y^2 : 'y' is missing This is simply a misunderstanding about

Re: [R] Default argument not passed to subfunction when argument name matches default expression

...@sciviews.org] Sent: Monday, March 31, 2014 9:40 AM To: R Help Subject: [R] Default argument not passed to subfunction when argument name matches default expression Hello, I have difficulties to understand this one: foo - function (y = 2) { bar - function (y = y) y^2 bar() } foo

Re: [R] Default argument not passed to subfunction when argument name matches default expression

Note that the fact that bar() is defined within foo() is irrelevant. ## At the top level/global prompt: y - 2 bar- function(y=y)y^2 bar() Error in y^2 : 'y' is missing ## but bar(y) [1] 4 This is due to lazy evaluation and promises: The formal argument y is not evaluated until it's used

Re: [R] Default argument not passed to subfunction when argument name matches default expression

On Mon, Mar 31, 2014 at 11:08 AM, Tierney, Luke luke-tier...@uiowa.edu wrote: The environment in which default arguments are evaluated is the environment of the function call itself, not the environment of the caller or the lexical enclosure (the same here). So the two 'y' used in function(y =