Sql 直接加法(1+1)与聚合函数SUM(1+1)的区别是什么

请参见下面的查询，其中我从没有函数的直接加法中获得了声誉值.

SELECT
    "User"."id",
    "User"."name",
    "User"."title",
    "User"."about",
    "User"."location",
    "User"."isModerator",
    "User"."createdAt",
    (
--      question vote
        COALESCE(SUM("PostVote"."value" * 10) FILTER (WHERE "value" = 1 AND "Post"."questionId" IS NULL), 0)::int 
        +
        COALESCE(SUM("PostVote"."value" * 2) FILTER (WHERE "value" = -1 AND "Post"."questionId" IS NULL), 0)::int 
        +
--      answer vote
        COALESCE(SUM("PostVote"."value" * 10) FILTER (WHERE "value" = 1 AND "Post"."questionId" IS NOT NULL), 0)::int 
        +
        COALESCE(SUM("PostVote"."value" * 2) FILTER (WHERE "value" = -1 AND "Post"."questionId" IS NOT NULL), 0)::int 
        +
--      accepted answer
        (COALESCE(COUNT("answer"."id"), 0)::int * 15)
    ) as reputation
FROM "User"
LEFT JOIN "Post" ON "Post"."authorId" = "User"."id"
LEFT JOIN "PostVote" ON "PostVote"."postId" = "Post"."id"
LEFT JOIN "Post" as answer ON "Post"."acceptedAnswerId" = "answer"."id"
GROUP BY "User"."id"
ORDER BY reputation DESC, id

我这样做是因为SUM不能嵌套，我最初想做这样的事情.

-- ...
    SUM(
--      question vote
        COALESCE(SUM("PostVote"."value" * 10) FILTER (WHERE "value" = 1 AND "Post"."questionId" IS NULL), 0)::int 
        +
        COALESCE(SUM("PostVote"."value" * 2) FILTER (WHERE "value" = -1 AND "Post"."questionId" IS NULL), 0)::int 
        +
--      answer vote
        COALESCE(SUM("PostVote"."value" * 10) FILTER (WHERE "value" = 1 AND "Post"."questionId" IS NOT NULL), 0)::int 
        +
        COALESCE(SUM("PostVote"."value" * 2) FILTER (WHERE "value" = -1 AND "Post"."questionId" IS NOT NULL), 0)::int 
        +
--      accepted answer
        (COALESCE(COUNT("answer"."id"), 0)::int * 15)
    ) as reputation
-- ...

我很好奇是否有任何性能影响或总结可能导致不正确的值？我已经用负值进行了测试，它工作得很好，也许我遗漏了一些东西.

我本可以通过嵌套查询来使用SUM函数，但我认为这太冗长了.