【框架设计】宠物技能书：技能槽解锁与顶替规则

---宠物技能槽：逐槽解锁的消耗
local SLOT_COST = { { 1, 0 }, { 2, 5 }, { 3, 20 }, { 4, 50 } }

function PetSlot.unlock(actor, slot)
    local player = class(actor)
    if not player then return false end
    -- 按槽位找消耗：第 2 槽 5 本技能书，第 4 槽 50 本
    local cost = 0
    for _, c in ipairs(SLOT_COST) do
        if c[1] == slot then cost = c[2] end
    end
    if not checkitemw(actor, "宠物技能书", cost) then
        sendmsg(actor, 1, "技能书不足，需 " .. cost .. " 本")
        return false
    end
    takeitem(actor, "宠物技能书", cost, 0, "技能槽解锁")
    setplayvar(actor, "HUMAN", "PetSlot_" .. slot, 1, 1)
    sendmsg(actor, 1, "技能槽 " .. slot .. " 已解锁")
    return true
end

---技能的顶替：确认与旧技能的处理
function PetSkill.learn(actor, slot, skillBookId)
    local player = class(actor)
    if not player then return false end
    local old = getplayvar(actor, 2,
        "PetSkill_" .. slot)
    if old and old ~= "" then
        -- 顶替需二次确认：旧技能将消失
        if getplayvar(actor, 2, "PetConfirm") ~= skillBookId then
            setplayvar(actor, "HUMAN", "PetConfirm",
                skillBookId, 0)
            sendmsg(actor, 1, "该槽位已有技能，再次使用将顶替")
            return false
        end
    end
    takeitem(actor, "技能书·" .. PetSkill.nameOf(skillBookId),
        1, 0, "宠物学习技能")
    setplayvar(actor, "HUMAN", "PetSkill_" .. slot,
        skillBookId, 1)
    PetSkill.apply(actor, slot, skillBookId)
    sendmsg(actor, 1, "宠物学会了新技能")
    return true
end