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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>
This commit is contained in:
Joseph Myers 2020-06-08 16:55:49 +00:00 committed by Paolo Bonzini
parent 6b8b0136ab
commit 499a2f7b55
1 changed files with 1 additions and 1 deletions
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2
fpu/softfloat.c
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@ -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);