https://gcc.gnu.org/bugzilla/show_bug.cgi?id=87390
--- Comment #6 from Vincent Lefèvre <vincent-gcc at vinc17 dot net> --- (In reply to jos...@codesourcery.com from comment #5) > It's 6.3.1.4 for conversions between real floating and integer types that, > in C99 but not C11, I think requires the resulting value to be > representable in the resulting real floating type. No, it just requires the value to be in the range of values that can be represented (which is always the case in practice, even though the C standard does not forbid very long integers). Anyway, this is unrelated to the use of extra precision and range. Note that: * 6.3.1.4 is about the behavior of conversion between real floating and integer, for given types. * 6.3.1.5 is about the behavior of conversion between real floating types, for given types. * 6.3.1.8 is about the selection of the arithmetic types (implicit conversions). Neither 6.3.1.4, nor 6.3.1.5 deals with semantic vs evaluation type (6.3.1.5 mentions "greater precision and range than required by its semantic type", but *only* for the other direction, in an *explicit* conversion). It is 6.3.1.8 that deals with the rules about extra precision and range. BTW, note 52 of 6.3.1.8 says "The cast and assignment operators are still required to perform their specified conversions as described in 6.3.1.4 and 6.3.1.5." i.e. mentioning both 6.3.1.4 and 6.3.1.5, thus confirming that implicit conversions of integer to real floating is also covered by the extra precision and range (otherwise referencing 6.3.1.4 would be pointless). That is, if d has type double, d + (1ULL << 63) and d + (double) (1ULL << 63) are handled differently.
