For-loop vs while loop in R

Question

I have noticed a curious thing whilst working in R. When I have a simple program that computes squares from 1 to N implemented using for-loop and while-loop the behaviour is not the same. (I don't care about vectorisation in this case or apply functions).

fn1 <- function (N) 
{
    for(i in 1:N) {
        y <- i*i
    }
}

AND

fn2 <- function (N) 
{
    i=1
    while(i <= N) {
        y <- i*i
        i <- i + 1
    }
}

The results are:

system.time(fn1(60000))
   user  system elapsed 
  2.500   0.012   2.493 
There were 50 or more warnings (use warnings() to see the first 50)
Warning messages:
1: In i * i : NAs produced by integer overflow
.
.
.

system.time(fn2(60000))
   user  system elapsed 
  0.138   0.000   0.137 

Now we know that for-loop is faster, my guess is because of pre allocation and optimisations there. But why does it overflow?

UPDATE: So now trying another way with vectors:

fn3 <- function (N) 
{
    i <- 1:N
    y <- i*i
}
system.time(fn3(60000))
   user  system elapsed 
  0.008   0.000   0.009 
Warning message:
In i * i : NAs produced by integer overflow

So Perhaps its a funky memory issue? I am running on OS X with 4Gb of memory and all default settings in R. This happens in 32- and 64-bit versions (except that times are faster).

Alex

Answer 1

Because 1 is numeric, but not integer (i.e. it's a floating point number), and 1:6000 is numeric and integer.

> print(class(1))
[1] "numeric"
> print(class(1:60000))
[1] "integer"

60000 squared is 3.6 billion, which is NOT representable in signed 32-bit integer, hence you get an overflow error:

> as.integer(60000)*as.integer(60000)
[1] NA
Warning message:
In as.integer(60000) * as.integer(60000) : NAs produced by integer overflow

3.6 billion is easily representable in floating point, however:

> as.single(60000)*as.single(60000)
[1] 3.6e+09

To fix your for code, convert to a floating point representation:

function (N)
{
    for(i in as.single(1:N)) {
        y <- i*i
    }
}