Re: Type length in bits?
"Alf P. Steinbach /Usenet" <alf.p.steinbach+usenet@gmail.com> schrieb im
Newsbeitrag news:ip0bud$639$1@dont-email.me...
* Angel Tsankov, on 24.04.2011 07:45:
What is a standard way to get the "length in bits" (in the sense of
5.8/1) of a type?
sizeof(T)*bitsPerByte
where 'bitsPerByte' can be defined as e.g.
#include <limits.h>
int const bitsPerByte = CHAR_BIT;
or
#include <limits>
int const bitsPerByte = std::numeric_limits<unsigned char>::digits;
The latter is both more verbose and more subtle.
Note that this bit length does not in general guarantee to tell you the
number of value representation bits, but mostly that's an academic, not a
practical, issue.
What happens if for a type T, the number of object representation bits
exceeds the number of value representation bits and I do this:
// Determine the number of bits used for the value representation.
const int num_bits = sizeof ( T ) * std::numeric_limits<T>::digits;
// The bit pattern of x will probably be all
// zeros except for the least significant bit.
T x = 1;
// Shift the least significant bit out of range of
// the bits used for the value representation.
x <<= num_bits;
Let us assume that T is a 18 bit integer type that uses 16 bits for the
value representation like this ( O = object representation only, V = value
representation):
OOVVVVVVVVVVVVVVVV
The first line will set the value of num_bits to 16. The second line will
initialize x to contain a value like ( U = undefined ):
UU0000000000000001
The thirs line will shift the least significant bit 16 positions to the
left, so it ends up at the position of the least significant bit used for
the object representation only:
U10000000000000000
Now what is the value of x according to the standard? Am I right in thinking
that it evaluates to zero, because any bits that are shifted out of range
are simply discarded? Or is it possible that the bit pattern now is a trap
representation on certain systems, resulting in undefined behaviour?
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