Function: bitprecision
Section: conversions
C-Name: bitprecision00
Prototype: GDG
Help: bitprecision(x,{n}): if n is present and positive, return x at precision
 n bits. If n is omitted, return real precision of object x in bits.
Doc: the function behaves differently according to whether $n$ is
 present or not. If $n$ is missing, the function returns
 the (floating point) precision in bits of the PARI object $x$.

 If $x$ is an exact object, the function returns \kbd{+oo}.
 \bprog
 ? bitprecision(exp(1e-100))
 %1 = 512                 \\ 512 bits
 ? bitprecision( [ exp(1e-100), 0.5 ] )
 %2 = 128                 \\ minimal accuracy among components
 ? bitprecision(2 + x)
 %3 = +oo                  \\ exact object
 @eprog\noindent Use \kbd{getlocalbitprec()} to retrieve the
 working bit precision (as modified by possible \kbd{localbitprec}
 statements).

 If $n$ is present and positive, the function creates a new object equal to $x$
 with the new bit-precision roughly $n$. In fact, the smallest multiple of 64
 (resp.~32 on a 32-bit machine) larger than or equal to $n$.

 For $x$ a vector or a matrix, the operation is
 done componentwise; for series and polynomials, the operation is done
 coefficientwise. For real $x$, $n$ is the number of desired significant
 \emph{bits}. If $n$ is smaller than the precision of $x$, $x$ is truncated,
 otherwise $x$ is extended with zeros. For exact or non-floating-point types,
 no change.
 \bprog
 ? bitprecision(Pi, 10)    \\ actually 64 bits ~ 19 decimal digits
 %1 = 3.141592653589793239
 ? bitprecision(1, 10)
 %2 = 1
 ? bitprecision(1 + O(x), 10)
 %3 = 1 + O(x)
 ? bitprecision(2 + O(3^5), 10)
 %4 = 2 + O(3^5)
 @eprog\noindent
