On Fri, 6 Sep 2013, Cong Hou wrote: > 4: (float) sqrtl ((long double) double_val) -> (float) sqrt (double_val)
I don't believe this case is in fact safe even if precision (long double) >= precision (double) * 2 + 2 (when your patch would allow it). The result that precision (double) * 2 + 2 is sufficient for the result of rounding the long double value to double to be the same as the result of rounding once from infinite precision to double would I think also mean the same when rounding of the infinite-precision result to float happens once - that is, if instead of (float) sqrt (double_val) you have fsqrt (double_val) (fsqrt being the proposed function in draft TS 18661-1 for computing a square root of a double value, rounded exactly once to float). But here the question isn't whether rounding to long double then float gives the same result as rounding to float. It's whether rounding to long double then float gives the same result as rounding to double then float. Consider, for example, double_val = 0x1.0000020000011p+0. The square root rounded once to float (and so the result if you go via long double and long double is sufficiently precise) is 0x1.000002p+0. But the square root rounded to double is 0x1.000001p+0, which on rounding to float produces 0x1p+0. > 5: (double) sqrtl ((long double) float_val) -> sqrt ((double) float_val) (This case, however, *is* safe if long double is sufficiently precise; the conversion of float to long double is trivially equivalent to a conversion involving an intermediate "double" type, so it reduces to a case for which the standard formula involving precisions of just two types applies.) -- Joseph S. Myers jos...@codesourcery.com
