iter33 卡② T2 变异伴生断言强化:V2d 补渲染面可见按钮计数锚(句柄缝盲区由变异A暴露——旧钮接回句柄外渲染不红,与卡③ sl4 同款教训);变异A(生涯钮接回→冒烟红)/变异B(进入存档接错层→屏流 5 红含 mn3)实测闭合
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@@ -238,6 +238,14 @@ func _topmost_visible_layer(main: Node) -> Node:
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return best
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# 渲染面可见按钮递归计数(V2d mn1/mn2 变异锚——句柄缝盲区闭合)
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func _count_visible_buttons(node: Node) -> int:
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var n := 1 if (node is Button and node.visible) else 0
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for c in node.get_children():
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n += _count_visible_buttons(c)
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return n
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# ---- boot 链路时序后置条件(启动态六断言;只立行为可判断言,不造假时序)----
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func _smoke_boot_chain() -> void:
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print("--- 迭代28 T2 boot 链路时序后置条件(V2:六断言)---")
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@@ -289,6 +297,8 @@ func _smoke_boot_chain() -> void:
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for old_btn in ["lvbtn", "resumebtn", "campbtn", "careerbtn"]:
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check(not (old_btn in main._menu_screen),
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"V2d 四旧钮句柄零命中·%s(彻底撤出——禁隐藏态禁置灰态残留)〔卡② mn2〕" % old_btn)
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check(_count_visible_buttons(main._start_overlay) == 3,
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"V2d 渲染面可见按钮计数==3(旧钮接回即红——渲染面锚,句柄缝盲区闭合)〔卡② mn1/mn2〕")
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# ⑤ 闲时会话(_ready 即建、永非 null、在树)
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check(main.run_scope != null and not main.run_scope.is_adventure and main.run_scope.is_inside_tree(),
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"V2e run_scope 非空且为闲时会话(is_adventure=false+is_inside_tree)〔装配生命批 V2〕")
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