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Supports customizing the display of all data types in Lua on the project side(支持在项目侧自定义Lua中所有类型数据的展示方式) #51

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merged 3 commits into from
Feb 19, 2024
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Supports customizing the display of all data types in Lua on the project side(支持在项目侧自定义Lua中所有类型数据的展示方式) #51

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15925a9
feat: Extend queryHelper condition to include all lua types
zhangjiequan Feb 6, 2024
e4bed6e
Revert "feat: Extend queryHelper condition to include all lua types"
zhangjiequan Feb 18, 2024
005e38a
feat:通过单纯修改调试器，实现项目侧自定义显示各种lua类型值的方式
zhangjiequan Feb 18, 2024
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4 changes: 3 additions & 1 deletion emmy_debugger/src/debugger/emmy_debugger.cpp
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Expand Up @@ -371,7 +371,9 @@ void Debugger::GetVariable(lua_State *L, Idx<Variable> variable, int index, int
variable->valueType = type;


if (queryHelper && (type == LUA_TTABLE || type == LUA_TUSERDATA || type == LUA_TFUNCTION)) {
if (queryHelper && (type == LUA_TTABLE || type == LUA_TUSERDATA || type == LUA_TFUNCTION
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应该改为类似 && registerType[type] (registerType应该是个数组) 的方式判断类型是否需要导出判断,

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你的意思是改为使用类似表驱动（或控制表, Control_table）的方式来实现？

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你加的这个功能对大部分人是不需要,所以影响尽可能的要小,判断类型是否在某个数组(不是set)上性能足够的高, 另外你需要提供函数: 注册需要导入到lua中的类型. 比如对外的接口类似dbg.registerType("string")

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你加的这个功能对大部分人是不需要,所以影响尽可能的要小,判断类型是否在某个数组(不是set)上性能足够的高, 另外你需要提供函数: 注册需要导入到lua中的类型. 比如对外的接口类似dbg.registerType("string")

话说，为什么使用数组而不是set？

有以下事实：

  1. std::unordered_set，基于哈希表，它的查找复杂度是O(1)
  2. std::vector，基于动态数组，它的查找复杂度是O(n)

为了高性能，是不是应该使用vector呢？
亦或，你的意思是，这里的元素量很小(n很小)，哈希算法反而更耗？

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@CppCXY CppCXY Feb 18, 2024 •
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std::array预先申请足够的内存(大概就8)然后把type的整型值作为索引去索引目标位置是否为true, 实际上更应该使用的是std::bitset, 或者直接使用整数的位运算

|| type == LUA_TSTRING || type == LUA_TNUMBER || type == LUA_TLIGHTUSERDATA || type == LUA_TTHREAD
|| type == LUA_TNONE || type == LUA_TNIL || type == LUA_TBOOLEAN)) {
if (manager->extension.QueryVariable(L, variable, typeName, index, depth)) {
return;
}
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