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我想知道是否可以在 R 中使用大于.Machine$double.xmax( ~1.79e308) 值的整数。我认为通过在 R 中使用 egRmpfr或gmp库，您可以分配任何大小的值，直到系统上的 RAM 限制？我认为这大于.Machine$double.xmax但显然不是。

> require( gmp )
> as.bigz( .Machine$double.xmax )
Big Integer ('bigz') :
[1] 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368
> as.bigz( 1e309 )
Big Integer ('bigz') :
[1] NA
>

有人可以解释为什么使用 64 位内存寻址的计算机不能存储大于 1.79e308 的值吗？抱歉 - 我没有计算机科学背景，但我正在努力学习。

谢谢。

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1 回答 1

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Rmpfr 可以使用mpfr_set_str进行字符串转换...

val <- mpfr("1e309")

## 1 'mpfr' number of precision  17   bits 
## [1] 9.999997e308

# set a precision (assume base 10)...
est_prec <- function(e) floor( e/log10(2) ) + 1

val <- mpfr("1e309", est_prec(309) )

## 1 'mpfr' number of precision  1027   bits 
## [1]1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

.mpfr2bigz(val)

## Big Integer ('bigz') :
## [1] 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

# extract exponent from a scientific notation string
get_exp <- function( sci ) as.numeric( gsub("^.*e",'', sci) )

# Put it together
sci2bigz <- function( str ) {
  .mpfr2bigz( mpfr( str, est_prec( get_exp( str ) ) ) )
}

val <- sci2bigz( paste0( format( Const("pi", 1027) ), "e309") )

identical( val, .mpfr2bigz( Const("pi",1027)*mpfr(10,1027)^309 ) )

## [1] TRUE

## Big Integer ('bigz') :
## [1] 3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587004

至于为什么要存储大于 的数字.Machine$double.xmax，IEEE 规范、R 常见问题解答和维基百科中有关浮点编码的文档进入了所有行话，但我发现仅定义术语（使用?'.Machine'）很有帮助...

double.xmax（最大归一化浮点数）=
(1 - double.neg.eps) * double.base ^ double.max.exp其中

  1. double.neg.eps（一个小的正浮点数 x 使得 1 - x != 1）=double.base ^ double.neg.ulp.digits其中
    • double.neg.ulp.digits= 最大的负整数，使得1 - double.base ^ i != 1和
  2. double.max.exp= double.base 溢出的最小正幂和
  3. double.base（浮点表示的基数）= 2（二进制）。

思考可以区分什么有限浮点数；IEEE 规范告诉我们，对于 binary64 数字，11 位用于指数，所以我们有一个最大指数，2^(11-1)-1=1023但我们想要溢出的最大指数，所以double.max.exp是 1024。

# Maximum number of representations
# double.base ^ double.max.exp
base <- mpfr(2, 2048)
max.exp <- mpfr( 1024, 2048 )

# This is where the big part of the 1.79... comes from
base^max.exp

## 1 'mpfr' number of precision  2048   bits 
## [1] 179769313486231590772930519078902473361797697894230657273430081157732675805500963132708477322407536021120113879871393357658789768814416622492847430639474124377767893424865485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684586298239947245938479716304835356329624224137216

# Smallest definitive unit.
# Find the largest negative integer...
neg.ulp.digits <- -64; while( ( 1 - 2^neg.ulp.digits ) == 1 ) 
  neg.ulp.digits <<- neg.ulp.digits + 1

neg.ulp.digits

## [1] -53

# It makes a real small number...
neg.eps <- base^neg.ulp.digits

neg.eps

## 1 'mpfr' number of precision  2048   bits 
## [1] 1.11022302462515654042363166809082031250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000e-16

# Largest difinitive floating point number less than 1
# times the number of representations
xmax <- (1-neg.eps) * base^max.exp

xmax

## 1 'mpfr' number of precision  2048   bits 
## [1] 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368

identical( asNumeric(xmax), .Machine$double.xmax )

## [1] TRUE
于 2015-06-15T23:04:46.273 回答