------- Comment #10 from fxcoudert at gcc dot gnu dot org 2008-03-01 12:08 ------- (In reply to comment #7) > I tried the attached change. It results in the correct configure > results. However, the test still fails.
By my reading, this patch will change nothing: it's OK for HAVE_BROKEN_ISFINITE not be defined if isfinite is not defined, because we then have: #if defined(HAVE_BROKEN_ISFINITE) || defined(__CYGWIN__) #undef isfinite #endif #if !defined(isfinite) // define isfinite #endif So, in your case, when isfinite and fpclassify are both functions (and not macros), we're going to simply define isfinite as: #define isfinite(x) ((x) - (x) == 0) David, when you have time, can you test the two following replacements: #define isnan(x) __builtin_expect ((x) != (x), 0) #define isfinite(x) __builtin_expect (!isnan((x) - (x)), 1) #define isinf(x) __builtin_expect (!isnan(x) & !isfinite(x), 0) and #define isnan(x) __builtin_isnan (x) #define isfinite(x) __builtin_isfinite (x) #define isinf(x) __builtin_isinf (x) (they should go in libgfortran/libgfortran.h and replace the following block of code: #if defined(HAVE_BROKEN_ISFINITE) || defined(__CYGWIN__) ... #define isnan(x) (fpclassify(x) == FP_NAN) #endif /* !defined(fpclassify) */ #endif /* !defined(isfinite) */ Thanks, FX -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=26252
