Date: Fri, 09 Apr 2021 21:34:31 +0000 From: bugzilla-noreply@freebsd.org To: bugs@FreeBSD.org Subject: =?UTF-8?B?W0J1ZyAyNTQ5MTFdIGxpYi9tc3VuL2N0cmlnX3Rlc3QgZmFpbHMg?= =?UTF-8?B?aWYgY29tcGlsZWQgd2l0aCBBVljCoCgtbWF2eCkgb3IgYW55IENQVVNFVCBl?= =?UTF-8?B?bmFibGluZyBBVlg=?= Message-ID: <bug-254911-227-Z1uWjjH16x@https.bugs.freebsd.org/bugzilla/> In-Reply-To: <bug-254911-227@https.bugs.freebsd.org/bugzilla/> References: <bug-254911-227@https.bugs.freebsd.org/bugzilla/>
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https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=3D254911 --- Comment #2 from Dimitry Andric <dim@FreeBSD.org> --- OK, for some reason the divide operation in the imaginary part in line 77 g= ives rise to FE_INVALID (0x1), but *only* if more advanced CPU instructions are used: 41 float complex 42 ctanhf(float complex z) 43 { ... 72 t =3D tanf(y); 73 beta =3D 1.0 + t * t; 74 s =3D sinhf(x); 75 rho =3D sqrtf(1 + s * s); 76 denom =3D 1 + beta * s * s; 77 return (CMPLXF((beta * rho * s) / denom, t / denom)); For example, one of the test cases has both x and y at zero, and this resul= ts in the values: t=3D0, beta=3D1, s=3D0, rho=3D1, denom=3D1 The real part of the return statement calculates (beta * rho * s) / denom = =3D (1 * 1 * 0) / 1 =3D 0, but this does *not* change the FP exception flags. The imaginary part calculates t / denom =3D 0 / 1, and for some reason this= sets FE_INVALID! It's tricky to debug this using printfs though, since this also causes FE_INEXACT, due to some floating point logic in the guts of printf. Now, as to why compiling for bog-standard x86_64, which *does* use SSE and SSE2, does not raise FE_INVALID, I'm not sure yet. --=20 You are receiving this mail because: You are the assignee for the bug.=
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