[Rd] bug in sum() on integer vector

[Uwe Ligges]

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