Puzzled over curve() syntax.

It's probably toadally elementary (and, like, duhhhhh) but
I can't figure out why the following doesn't work:

	curve(function(x){qnorm(x,4,25)},from=0,to=1)

I get the error:

	Error in xy.coords(x, y, xlabel, ylabel, log) : 
        'x' and 'y' lengths differ

But if I do

	foo <- function(x){qnorm(x,4,25)}
	curve(foo,from=0,to=1)

it goes like a train.

Also

	plot(function(x){qnorm(x,4,25)},from=0,to=1)

works just fine.

I'm using

 > version

         _                   
platform sparc-sun-solaris2.9
arch     sparc               
os       solaris2.9          
system   sparc, solaris2.9   
status                       
major    2                   
minor    2.0                 
year     2005                
month    10                  
day      06                  
svn rev  35749               
language R 

This is just idle curiousity I guess, but I would like to deepen my
understanding.  There's probably something about the ``expression''
concept that I'm not grokking here ....

It's the way curve is written (and documented, though perhaps a little 
obscurely).  If you debug it, you'll see that eventually your function 
gets assigned to a variable called "expr", and a nice list of values 
gets assigned to "x", then it tries to evaluate

  y <- eval(expr, envir = list(x = x), enclos = parent.frame())


But if you evaluate expr, you just get the function back, you don't call 
it.  The problem is that curve was written assuming you'd call it as

curve(qnorm(x,4,25),from=0,to=1)

in which case the expression "qnorm(x,4,25)" gets evaluated at those x 
values and things are fine.

I don't think this is a bug, but it might be worth fixing so your code 
works too.  It's a little tricky, because to know that you passed a 
function in, you probably want to evaluate it; but if you evaluate 
"qnorm(x,4,25)" before you've set up x, you'll get an error.

A fix is to add an additional else clause after the first test, namely

     else if (is.language(sexpr) && identical(sexpr[[1]], 
as.name("function"))) {
           expr <- substitute(do.call(expr, list(x)), list(expr=expr))
           if (is.null(ylab))
               ylab <- deparse(sexpr)
     }

but this still doesn't handle the case where you've given a more general 
expression that returns a function, e.g. picking one out of a list. 
You'll probably need another argument to distinguish the case of an 
expression returning y values from an expression returning a function, 
and I'm not sure that level of elaboration would really be a good idea.

Duncan Murdoch