Comments (3)
ben620
commented on May 19, 2024
lua中调用一个不存在的函数，工具不会给出提示。
目前不能 。工具区分了函数和变量。重点处理了变量的问题。
因为lua组要和其他语言配合使用，lua中未定义的函数不见得函数是没有定义的。
统计了很多项目的使用情况，才做出这种策略
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kona8
commented on May 19, 2024
这款工具可以发现许多代码问题，效果挺好的。但是对table的检测似乎不是很强大，比如对table中不存在的成员的使用不会报错，经常由于笔误写错检测不出来。
另外对于lua_SingleScopeGlobal选项有些疑惑，在a.lua中定义一个全局table和成员，在b.lua中使用，检测会不能通过。
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ben620
commented on May 19, 2024
table的检查有一条规则。检测能力比较弱。
另外一条table相关检查规则再内部试用了，后续会对外更新。
lua_SingleScopeGlobal这个规则实现有些问题，方便的话可以提供协议代码示例，帮助我们改进
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