Assignment using call()

Question

Suppose I want to manipulate function body like this:

> f1 = function(a, b, d) {
+   stop("This is a template!")
+   result
+ }
> body(f1)[[2]] = call("<-", as.name("result"), lapply(letters[1:2], as.name))

This looks good...

> f1
function (a, b, d) 
{
    result <- list(a, b)
    result
}

...but doesn't work:

> f1(a = 123, b = 456, d = 999)
[[1]]
a

[[2]]
b

On the other hand, when I do this:

> body(f1)[[2]] = call("<-", as.name("result"),
+                      as.call(c(as.name("list"), lapply(letters[1:2], as.name))))

It looks the same...

> f1
function (a, b, d) 
{
    result <- list(a, b)
    result
}

...but it works:

> f1(a = 123, b = 456, d = 999)
[[1]]
[1] 123

[[2]]
[1] 456

Can somebody please break this down for me into some really small pieces and explain what exactly is happening here?

Answer 1

The problem is that deparse is not perfect. Incidentally this is unrelated to the assignment; the same problem exists when the expression is used in a different context, e.g.:

bquote(1 + .(lapply(letters[1:2], as.name)))

or

bquote(sum(.(lapply(letters[1:2], as.name))))

Either way, you’re constructing a list of names, but you are not constructing a call to list inside an unevaluated expression. But (I’m guessing that) the R deparser doesn’t know what to do with a list of names in an expression, since that situation can’t occur in real R code that isn’t constructed via unevalued expressions.