softfloat: fix floatx80 remainder pseudo-denormal check for zero

The floatx80 remainder implementation ignores the high bit of the
significand when checking whether an operand (numerator) with zero
exponent is zero.  This means it mishandles a pseudo-denormal
representation of 0x1p-16382L by treating it as zero.  Fix this by
checking the whole significand instead.

Signed-off-by: Joseph Myers <joseph@codesourcery.com>
Reviewed-by: Richard Henderson <richard.henderson@linaro.org>
Message-Id: <alpine.DEB.2.21.2006081655180.23637@digraph.polyomino.org.uk>
Signed-off-by: Paolo Bonzini <pbonzini@redhat.com>

diff --git a/fpu/softfloat.c b/fpu/softfloat.c
index b7dcf4d6c39989667a2f53c6d2029f2000cae95a..f164b5c0adb246d972b35bc6f9ee9cbc1c48435d 100644 (file)
--- a/fpu/softfloat.c
+++ b/fpu/softfloat.c
@@ -5741,7 +5741,7 @@ floatx80 floatx80_modrem(floatx80 a, floatx80 b, bool mod,
         normalizeFloatx80Subnormal( bSig, &bExp, &bSig );
     }
     if ( aExp == 0 ) {
-        if ( (uint64_t) ( aSig0<<1 ) == 0 ) return a;
+        if ( aSig0 == 0 ) return a;
         normalizeFloatx80Subnormal( aSig0, &aExp, &aSig0 );
     }
     bSig |= UINT64_C(0x8000000000000000);