[Rd] Reading 64-bit integers

Jon Clayden jon.clayden at gmail.com
Wed Mar 30 01:01:17 CEST 2011


Dear Simon,

On 29 March 2011 22:40, Simon Urbanek <simon.urbanek at r-project.org> wrote:
> Jon,
>
> On Mar 29, 2011, at 1:33 PM, Jon Clayden wrote:
>
>> Dear Simon,
>>
>> Thank you for the response.
>>
>> On 29 March 2011 15:06, Simon Urbanek <simon.urbanek at r-project.org> wrote:
>>>
>>> On Mar 29, 2011, at 8:46 AM, Jon Clayden wrote:
>>>
>>>> Dear all,
>>>>
>>>> I see from some previous threads that support for 64-bit integers in R
>>>> may be an aim for future versions, but in the meantime I'm wondering
>>>> whether it is possible to read in integers of greater than 32 bits at
>>>> all. Judging from ?readBin, it should be possible to read 8-byte
>>>> integers to some degree, but it is clearly limited in practice by R's
>>>> internally 32-bit integer type:
>>>>
>>>>> x <- as.raw(c(0,0,0,0,1,0,0,0))
>>>>> (readBin(x,"integer",n=1,size=8,signed=F,endian="big"))
>>>> [1] 16777216
>>>>> x <- as.raw(c(0,0,0,1,0,0,0,0))
>>>>> (readBin(x,"integer",n=1,size=8,signed=F,endian="big"))
>>>> [1] 0
>>>>
>>>> For values that fit into 32 bits it works fine, but for larger values
>>>> it fails. (I'm a bit surprised by the zero - should the value not be
>>>> NA if it is out of range?
>>>
>>> No, it's not out of range - int is only 4 bytes so only 4 first bytes (respecting endianness order, hence LSB) are used.
>>
>> The fact remains that I ask for the value of an 8-byte integer and
>> don't get it.
>
> I think you're misinterpreting the documentation:
>
>     If ‘size’ is specified and not the natural size of the object,
>     each element of the vector is coerced to an appropriate type
>     before being written or as it is read.
>
> The "integer" object type is defined as signed 32-bit in R, so if you ask for "8 bytes into object type integer", you get a coercion into that object type -- 32-bit signed integer -- as documented. I think the issue may come from the confusion of the object type "integer" with general "integer number" in mathematical sense that has no representation restrictions. (FWIW in C the "integer" type is "int" and it is 32-bit on all modern OSes regardless of platform - that's where the limitation comes from, it's not something R has made up).

OK, but it still seems like there is a case for raising a warning. As
it is there is no way to tell when reading an 8-byte integer from a
file whether its value is really 0, or if it merely has 0 in its
least-significant 4 bytes. If 99% of such stored numbers are below
2^31, one is going to need some extra logic to catch the other 1%
where you (silently) get the wrong value. In essence, unless you're
certain that you will never come across a number that actually uses
the upper 4 bytes, you will always have to read it as two 4-byte
numbers and check that the high-order one (which is endianness
dependent, of course) is zero. A C-level sanity check seems more
efficient and more helpful to me.

>> Pretending that it's really only four bytes because of
>> the limits of R's integer type isn't all that helpful. Perhaps a
>> warning should be put out if the cast will affect the value of the
>> result? It looks like the relevant lines in src/main/connections.c are
>> 3689-3697 in the current alpha:
>>
>> #if SIZEOF_LONG == 8
>>                   case sizeof(long):
>>                       INTEGER(ans)[i] = (int)*((long *)buf);
>>                       break;
>> #elif SIZEOF_LONG_LONG == 8
>>                   case sizeof(_lli_t):
>>                       INTEGER(ans)[i] = (int)*((_lli_t *)buf);
>>                       break;
>> #endif
>>
>>>> ) The value can be represented as a double,
>>>> though:
>>>>
>>>>> 4294967296
>>>> [1] 4294967296
>>>>
>>>> I wouldn't expect readBin() to return a double if an integer was
>>>> requested, but is there any way to get the correct value out of it?
>>>
>>> Trivially (for your unsigned big-endian case):
>>>
>>> y <- readBin(x, "integer", n=length(x)/4L, endian="big")
>>> y <- ifelse(y < 0, 2^32 + y, y)
>>> i <- seq(1,length(y),2)
>>> y <- y[i] * 2^32 + y[i + 1L]
>>
>> Thanks for the code, but I'm not sure I would call that trivial,
>> especially if one needs to cater for little endian and signed cases as
>> well!
>
> I was saying for your case and it's trivial as in read as integers, convert to double precision and add.
>
>
>> This is what I meant by reconstructing the number manually...
>>
>
> You didn't say so - you were talking about reconstructing it from a raw vector which seems a lot more painful since you can't compute with enough precision on raw vectors.

True - I should have been more specific. Sorry.

Jon



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