From c69d0d08d0d71c779a245babe80342f0cf1ea985 Mon Sep 17 00:00:00 2001
From: Gael Guennebaud <g.gael@free.fr>
Date: Mon, 18 Feb 2019 14:43:07 +0100
Subject: Set cost of conjugate to 0 (in practice it boils down to a no-op).
 This is also important to make sure that A.conjugate() * B.conjugate() does
 not evaluate its arguments into temporaries (e.g., if A and B are fixed and
 small, or * fall back to lazyProduct)

---
 Eigen/src/Core/functors/UnaryFunctors.h | 10 +++++++++-
 1 file changed, 9 insertions(+), 1 deletion(-)

(limited to 'Eigen/src/Core/functors')

diff --git a/Eigen/src/Core/functors/UnaryFunctors.h b/Eigen/src/Core/functors/UnaryFunctors.h
index 55994047e..1d5eb3678 100644
--- a/Eigen/src/Core/functors/UnaryFunctors.h
+++ b/Eigen/src/Core/functors/UnaryFunctors.h
@@ -117,7 +117,15 @@ template<typename Scalar>
 struct functor_traits<scalar_conjugate_op<Scalar> >
 {
   enum {
-    Cost = NumTraits<Scalar>::IsComplex ? NumTraits<Scalar>::AddCost : 0,
+    Cost = 0,
+    // Yes the cost is zero even for complexes because in most cases for which
+    // the cost is used, conjugation turns to be a no-op. Some examples:
+    //   cost(a*conj(b)) == cost(a*b)
+    //   cost(a+conj(b)) == cost(a+b)
+    //   <etc.
+    // If we don't set it to zero, then:
+    //   A.conjugate().lazyProduct(B.conjugate())
+    // will bake its operands. We definitely don't want that!
     PacketAccess = packet_traits<Scalar>::HasConj
   };
 };
-- 
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