R-alpha: lazy evaluation and plot.step()

f<-function(a,b){a<-1; print(deparse(substitute(c(a,b))))}
f(a,b)

f<-function(a,b){a<-1; print(deparse(substitute(c(a,b))))}
f(a,b)

f(2,b)

f(2:2,b)

f("a",b)    

"test" <-
function(x, y, xlab = deparse(substitute(x)), ylab = deparse(substitute(y)))
{
  x <- sort(x)
  print(xlab)
  n <- length(x)
  y <- (1:n) / n
  print(ylab)
}

Under R,

R> test(4:1)
[1] "c(1, 2, 3, 4)"
[1] "c(0.25, 0.5, 0.75, 1)"
R> test(4:1, 4)
[1] "c(1, 2, 3, 4)"
[1] "c(0.25, 0.5, 0.75, 1)"

Under S,

test(4:1)

[1] "4:1"
[1] ""
[1] ""

test(4:1, 4)

[1] "4:1"
[1] "c(0.25, 0.5, 0.75, 1)"
[1] "c(0.25, 0.5, 0.75, 1)"

???

Peter Dalgaard BSA writes:

The explanation seems to be in different behaviour of the
substitute(...) construction:

R:

f<-function(a,b){a<-1; print(deparse(substitute(c(a,b))))}
f(a,b)

[1] "c(1, b)"

Splus:

f<-function(a,b){a<-1; print(deparse(substitute(c(a,b))))}
f(a,b)

[1] "c(a, b)"
[1] "c(a, b)"

So in R, assignment to a formal parameter causes the changed value to
be used for substitution purposes.

In S, this does not happen *unless the formal parameter is a
constant.*!

f(2,b)

[1] "c(1, b)"
[1] "c(1, b)"

f(2:2,b)

[1] "c(2:2, b)"
[1] "c(2:2, b)"

f("a",b)    

[1] "c(1, b)"
[1] "c(1, b)"

This is confusing in both languages, but I'd say that S has the bigger
problem...

In Kurts example, inserting 

xlab 
ylab

as the first two lines of the function defeats lazy evaluation and
provides the desired behaviour (I suppose) in both R and S.