Mike Stump <mikest...@comcast.net> writes: >>> * A static assert also prevents fixed_wide_ints from being initialised >>> from wide_ints. I think combinations like that would always be a >>> mistake. > > I don't see why. In C, this: > > int i; > long l = i; > > is not an error, and while a user might not want to do this, other > times, it is completely reasonable.
Sure, but in these terms, "int" is FLEXIBLE_PRECISION, while wide_int is VAR_PRECISION. You can do the above because "int" has a sign, and so the compiler knows how to extend it to wider types. wide_int doesn't have a sign, so if you want to extend it to wider types you need to specify a sign explicitly. So you can do: max_wide_int x = max_wide_int::from (wide_int_var, SIGNED); max_wide_int x = max_wide_int::from (wide_int_var, UNSIGNED); What I'm saying is that the assert stops plain: max_wide_int x = wide_int_var; since it's very unlikely that you'd know up-front that wide_int_var has precision MAX_BITSIZE_MODE_ANY_INT. FLEXIBLE_PRECISION is for integers that act in a C-like way. CONST_PRECISION and VAR_PRECISION are (deliberately) not C-like. Thanks, Richard
