CODEWITHSUNDEEP
rfunctionarguments
g g
g g

Reputation: 87

How is R passing parameters here?

I'm not that much of an OOP guy, so could someone please explain this simple concept in layman terms.

When I call foo.child from the foo.parent function, it seems to pass the A=7 argument down into the foo.child object and overrides or takes precedence over the A=3 default argument in foo.child as I would expect. 

foo.parent <- function(A=7) foo.child(A)
foo.child <- function(A=3) 2+A
foo.parent(A=7)
#[1] 9

But when I instantiate it inside of foo.parent, the parameter A=7 does pass down or force the instantiated object to use A=7; instead it uses the child object's parameter of A=3

foo.child<-function(A=3) 2+A
foo.parent <- function(A=7){
  foo.child(A=3)
}
foo.parent(A=7)
#[1] 5

Why does that happen? And what terminology would I use to describe the differences?

Upvotes: 0

Views: 373

Answers (2)

csgillespie
csgillespie

Reputation: 60492

I think your problem is you don't quite understand how default arguments work. So

foo.child = function(A=1) 2+A

defines the function foo.child that has a default argument A=1. So,

foo.child()

gives 3. Now in this function

foo.parent = function(A=3){
   foo.child(A=4)
}

you are always passing the value A=4 to the function foo.child, hence,

foo.parent(A=7)
# 6

Also, when you are trying to figure out what is happening, it's helpful to have different values of A

Upvotes: 2

Daniel Fischer
Daniel Fischer

Reputation: 3380

In your second example you do not give a value to A (At least not in such a way as you might thought). Try

foo.child<-function(A=3) 2+A
foo.parent<-function(A=7){
     foo.child(A=A)
}
foo.parent(A=7)
foo.parent()

instead. So, you mix here two different As. The =sign within a function call defines, what happens if you do not give a value for that variable in the function call.

Upvotes: 2

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