Endians and c++

[Kalle Rutanen]

Hi

Is it endian-safe to use bit operations in C++ ?

For example bit-shifts, does the left shift by n always equal
multiplicication by 2^n (modulo 2^integerBits), independent of endian
system used ?

[Victor Bazarov]

Kalle Rutanen wrote:

Is it endian-safe to use bit operations in C++ ?

For example bit-shifts, does the left shift by n always equal
multiplicication by 2^n (modulo 2^integerBits), independent of endian
system used ?

Yes. The bit shifts operate on the number _value_, the representation
of the number in memory has nothing to do with it.

V

[msalters]

Kalle Rutanen wrote:

Hi

Is it endian-safe to use bit operations in C++ ?

For example bit-shifts, does the left shift by n always equal
multiplicication by 2^n (modulo 2^integerBits), independent of endian
system used ?

No, but that's actually unrelated to endianness. Multiplication by
2^-1 is well defined, but a left shift over -1 bits isn't.
Overflow is undefined for signed types.

For unsigned types and positive n, the result is in fact defined as if
done by multiplication or division. (5.8 Shift operations)

HTH,
Michiel Salters

[Victor Bazarov]

msalters wrote:

Kalle Rutanen wrote:

Hi

Is it endian-safe to use bit operations in C++ ?

For example bit-shifts, does the left shift by n always equal
multiplicication by 2^n (modulo 2^integerBits), independent of endian
system used ?

No, but that's actually unrelated to endianness. Multiplication by
2^-1 is well defined, but a left shift over -1 bits isn't.

Wouldn't left shift by a negative number be the same as right shift by
the absolute value of the number? I mean, theoretically, of course.
[...]

V

[Kalle Rutanen]

Thanks for both repliers!

[msalters]

Victor Bazarov wrote:

msalters wrote:

Kalle Rutanen wrote:

Hi

Is it endian-safe to use bit operations in C++ ?

For example bit-shifts, does the left shift by n always equal
multiplicication by 2^n (modulo 2^integerBits), independent of endiansystem used ?

No, but that's actually unrelated to endianness. Multiplication by
2^-1 is well defined, but a left shift over -1 bits isn't.

Wouldn't left shift by a negative number be the same as right shift

by the absolute value of the number? I mean, theoretically, of course.

Yes, of course. If the shift operand is a literal or a constant, this
can
be an automatic transformation. However, if the rhs is a variable, this
cannot be done at compile-time. Now, apparently some CPU's dont't like
negative arguments, and it would be bad if they have have to implement
every << as (IF RHS<0 (NEG RHS, SHR RHS) ELSE SHL RHS ). This can be
done in C++ if needed, but you don't pay if you don't need it.

HTH,
Michiel Salters