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[24.113.145.216]) by smtp.gmail.com with ESMTPSA id z11sm285852pfk.46.2020.07.08.09.07.41 (version=TLS1_3 cipher=TLS_AES_128_GCM_SHA256 bits=128/128); Wed, 08 Jul 2020 09:07:41 -0700 (PDT) Subject: Re: [PATCH 03/21] softfloat: add xtensa specialization for pickNaNMulAdd To: Max Filippov , qemu-devel@nongnu.org References: <20200706234737.32378-1-jcmvbkbc@gmail.com> <20200706234737.32378-4-jcmvbkbc@gmail.com> From: Richard Henderson Message-ID: Date: Wed, 8 Jul 2020 09:07:39 -0700 User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:68.0) Gecko/20100101 Thunderbird/68.8.0 MIME-Version: 1.0 In-Reply-To: <20200706234737.32378-4-jcmvbkbc@gmail.com> Content-Type: text/plain; charset=utf-8 Content-Language: en-US Content-Transfer-Encoding: 7bit Received-SPF: pass client-ip=2607:f8b0:4864:20::1041; envelope-from=richard.henderson@linaro.org; helo=mail-pj1-x1041.google.com X-detected-operating-system: by eggs.gnu.org: No matching host in p0f cache. That's all we know. X-Spam_score_int: -20 X-Spam_score: -2.1 X-Spam_bar: -- X-Spam_report: (-2.1 / 5.0 requ) BAYES_00=-1.9, DKIM_SIGNED=0.1, DKIM_VALID=-0.1, DKIM_VALID_AU=-0.1, DKIM_VALID_EF=-0.1, RCVD_IN_DNSWL_NONE=-0.0001, SPF_HELO_NONE=0.001, SPF_PASS=-0.001 autolearn=ham autolearn_force=no X-Spam_action: no action X-BeenThere: qemu-devel@nongnu.org X-Mailman-Version: 2.1.23 Precedence: list List-Id: List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Cc: Peter Maydell , =?UTF-8?Q?Alex_Benn=c3=a9e?= Errors-To: qemu-devel-bounces+qemu-devel=archiver.kernel.org@nongnu.org Sender: "Qemu-devel" On 7/6/20 4:47 PM, Max Filippov wrote: > pickNaNMulAdd logic on Xtensa is the same as pickNaN when applied to > the expression (a * b) + c. So with two pickNaN variants there must be > two pickNaNMulAdd variants. "Is the same as"? I question the non-use of the infzero parameter. When infzero, (a * b) = (Inf * 0), which will produce a default QNaN. Your sentence above would suggest that pickNaN is applied twice, so that if use_first_nan, the default nan is chosen above any nan within c. In addition, is the invalid flag raised for (Inf * 0) + NaN? Does that happen regardless of the use_first_nan setting, or does the whole operation short-circuit? r~