Hi,
I believe it's lazy evaluation. See ?force
HTH,
baptiste
On 14 February 2010 20:32, Jyotirmoy Bhattacharya
jyotir...@jyotirmoy.net wrote:
I want to use lapply and a function returning a function in order to build a
list of functions.
genr1 - function(k) {function() {k}}
l1 - lapply(1:2,genr1)
l1[[1]]()
[1] 2
This was unexpected. I had expected the answer to be 1, since that is the
value k should be bound to when genr1 is applied to the first element of
1:2.
By itself genr1 seems to work fine.
genr1(5)()
[1] 5
I defined a slightly different higher-order function:
genr2 - function(k) {k;function() {k}}
l2 - lapply(1:2,genr2)
l2[[1]]()
[1] 1
This gives the answer I expected.
Now I am confused. The function returned by genr2 is exactly the same
function that was being returned by genr1. Why should evaluating k make a
difference?
I am using R 2.9.2 on Ubuntu Linux.
Jyotirmoy Bhattacharya
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