Re: [Rd] bug in sum() on integer vector

From: John C Nash <nashjc_at_uottawa.ca>
Date: Wed, 14 Dec 2011 16:16:31 -0500

I agree that where the overflow occurs is not critical (one can go back to cumsum and find out). I am assuming that Uwe still wants to know there has been an overflow at some point i.e., a warning. This could become more "interesting" as parallel computation causes different summation orderings on sums of large numbers of items.

JN

On 12/14/2011 03:58 PM, Uwe Ligges wrote:

> 
> 
> On 14.12.2011 17:19, peter dalgaard wrote:

>>
>> On Dec 14, 2011, at 16:19 , John C Nash wrote:
>>
>>>
>>> Following this thread, I wondered why nobody tried cumsum to see where the integer
>>> overflow occurs. On the shorter xx vector in the little script below I get a message:
>>>
>>> Warning message:
>>> Integer overflow in 'cumsum'; use 'cumsum(as.numeric(.))'
>>>>
>>>
>>> But sum() does not give such a warning, which I believe is the point of contention. Since
>>> cumsum() does manage to give such a warning, and show where the overflow occurs, should
>>> sum() not be able to do so? For the record, I don't class the non-zero answer as an error
>>> in itself. I regard the failure to warn as the issue.
>>
>> It (sum) does warn if you take the two "halves" separately. The issue is that the
>> overflow is detected at the end of the summation, when the result is to be saved to an
>> integer (which of course happens for all intermediate sums in cumsum)
>>
>>> x<- c(rep(1800000003L, 10000000), -rep(1200000002L, 15000000))
>>> sum(x[1:10000000])
>> [1] NA
>> Warning message:
>> In sum(x[1:1e+07]) : Integer overflow - use sum(as.numeric(.))
>>> sum(x[10000001:25000000])
>> [1] NA
>> Warning message:
>> In sum(x[10000001:1.5e+07]) : Integer overflow - use sum(as.numeric(.))
>>> sum(x)
>> [1] 4996000
>>
>> There's a pretty easy fix, essentially to move
>>
>> if(s> INT_MAX || s< R_INT_MIN){
>> warningcall(call, _("Integer overflow - use sum(as.numeric(.))"));
>> *value = NA_INTEGER;
>> }
>>
>> inside the summation loop. Obviously, there's a speed penalty from two FP comparisons
>> per element, but I wouldn't know whether it matters in practice for anyone.
>>
> 
> 
> I don't think I am interested in where the overflow happens if I call sum()...
> 
> Uwe

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