makes implementation of Animation::operator =(const Animation&) concise

Using the logical AND, we can evaluate all the expressions
from left to right into a single and condensed expression.

Change-Id: I973061dcd34322c29565a478bf9b7ba757965231
Signed-off-by: Adrien Ollier <adr.ollier@hotmail.fr>
Reviewed-on: https://gerrit.libreoffice.org/71651
Tested-by: Jenkins
Reviewed-by: Mike Kaganski <mike.kaganski@collabora.com>

vcl/source/animate/Animation.cxx

@@ -86,25 +86,13 @@ Animation& Animation::operator=(const Animation& rAnimation)
 
 bool Animation::operator==(const Animation& rAnimation) const
 {
-    const size_t nCount = maList.size();
-    bool bRet = false;
-
-    if (rAnimation.maList.size() == nCount && rAnimation.maBitmapEx == maBitmapEx
-        && rAnimation.maGlobalSize == maGlobalSize)
-    {
-        bRet = true;
-
-        for (size_t n = 0; n < nCount; n++)
-        {
-            if ((*maList[n]) != (*rAnimation.maList[n]))
-            {
-                bRet = false;
-                break;
-            }
-        }
-    }
-
-    return bRet;
+    return maList.size() == rAnimation.maList.size() && maBitmapEx == rAnimation.maBitmapEx
+           && maGlobalSize == rAnimation.maGlobalSize
+           && std::equal(maList.begin(), maList.end(), rAnimation.maList.begin(),
+                         [](const std::unique_ptr<AnimationBitmap>& pAnim1,
+                            const std::unique_ptr<AnimationBitmap>& pAnim2) -> bool {
+                             return *pAnim1 == *pAnim2;
+                         });
 }
 
 void Animation::Clear()