Is it legal to static_assert that signed shift right has two's-complement behavior?

Question

Is it legal to do the following in C11, C++11 and C++14?

static_assert(((-4) >> 1) == -2, "my code assumes sign-extending right shift");

or the C equivalent:

_Static_assert(((-4) >> 1) == -2, "my code assumes sign-extending right shift");

I don't know the rules for constant-expressions regarding whether you can use implementation-defined operations like the above.

I'm aware that the opposite, signed shift left of negative numbers, is undefined regardless of machine type.

Answer 1

Yes. The C++11 standard says in [expr.shift]/3:

The value of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a non-negative value, the value of the result is the integral part of the quotient of E1/2^E2 . If E1 has a signed type and a negative value, the resulting value is implementation-defined.

And nowhere in [expr.const]/2 it is said that such a shift, or expressions with implementation-defined values in general, are not constant expressions. You will thus get a constant expression that has an implementation-defined value.

Answer 2

This is legal, insofaras it doesn't cause undefined behaviour.

The behaviour of right-shift of negative values is implementation-defined . The C and C++ standards do not guarantee it to be either arithmetic or logical; although so far as I know there has never been a CPU that didn't pick one or the other.