英文:
Saving the output from Lua in C++ with SOL3 to an std::string
问题
我正在尝试将Lua解释器实现到我的C++代码中。我已经使用ImGui为我的项目实现了一个小型编辑器，并将编辑器的输出保存到一个std::vector<char>中。
我的Lua解释器的尝试实现如下：
// 头文件
std::string ExecuteLua();
std::vector<char> m_luaEditorData;
// cpp 文件
std::string Proxy::ExecuteLua()
{
// 从字符串加载Lua代码
std::string luaCode(m_luaEditorData.data());
// 创建Lua状态
sol::state lua;
// 加载标准Lua库
lua.open_libraries(sol::lib::base, sol::lib::package, sol::lib::string, sol::lib::table);
// 执行Lua代码并存储结果
sol::protected_function_result result = lua.script(luaCode);
// 检查错误
if (!result.valid())
{
sol::error error = result;
std::string errorMsg = error.what();
return "Lua错误: " + errorMsg;
}
// 将结果作为字符串获取
std::string output = lua["tostring"](result.get<sol::object>());
// 返回输出
return output;
}
...
if (m_luaEditorData.empty())
m_luaEditorData.push_back('// 头文件
std::string ExecuteLua();
std::vector<char> m_luaEditorData;
// cpp 文件
std::string Proxy::ExecuteLua()
{
// 从字符串加载Lua代码
std::string luaCode(m_luaEditorData.data());
// 创建Lua状态
sol::state lua;
// 加载标准Lua库
lua.open_libraries(sol::lib::base, sol::lib::package, sol::lib::string, sol::lib::table);
// 执行Lua代码并存储结果
sol::protected_function_result result = lua.script(luaCode);
// 检查错误
if (!result.valid())
{
sol::error error = result;
std::string errorMsg = error.what();
return "Lua错误: " + errorMsg;
}
// 将结果作为字符串获取
std::string output = lua["tostring"](result.get<sol::object>());
// 返回输出
return output;
}
...
if (m_luaEditorData.empty())
m_luaEditorData.push_back('\0');
auto luaEditorFlags = ImGuiInputTextFlags_AllowTabInput | ImGuiInputTextFlags_CallbackResize;
ImGui::InputTextMultiline("##LuaEditor", m_luaEditorData.data(), m_luaEditorData.size(), ImVec2(ImGui::GetWindowContentRegionWidth(), ImGui::GetWindowHeight() - (ImGui::GetTextLineHeight() * 16), luaEditorFlags, ResizeInputTextCallback, &m_luaEditorData);
...
');
auto luaEditorFlags = ImGuiInputTextFlags_AllowTabInput | ImGuiInputTextFlags_CallbackResize;
ImGui::InputTextMultiline("##LuaEditor", m_luaEditorData.data(), m_luaEditorData.size(), ImVec2(ImGui::GetWindowContentRegionWidth(), ImGui::GetWindowHeight() - (ImGui::GetTextLineHeight() * 16), luaEditorFlags, ResizeInputTextCallback, &m_luaEditorData);
...
当我运行这段代码时，我的输出中只会得到nil，正确的输出应该是标准输出（我不希望它输出在这里，而是输出到我的std::string中），当我输入错误的代码时，它会在sol.hpp中抛出异常。我没有找到如何执行此操作的示例，因此我正在尝试自己解决这个问题。
英文:
I'm trying to implement a lua interpreter to my C++ code. I have implemented a small editor for my project using ImGui and I'm saving the output from the editor to an std::vector<char>.
My attempted implementation of my lua interpeter looks like so;
// header
std::string ExecuteLua();
std::vector<char> m_luaEditorData;
// cpp
std::string Proxy::ExecuteLua()
{
// Load the Lua code from the string
std::string luaCode(m_luaEditorData.data());
// Create a Lua state
sol::state lua;
// Load standard Lua libraries
lua.open_libraries(sol::lib::base, sol::lib::package, sol::lib::string, sol::lib::table);
// Execute the Lua code and store the result
sol::protected_function_result result = lua.script(luaCode);
// Check for errors
if (!result.valid())
{
sol::error error = result;
std::string errorMsg = error.what();
return "Lua error: " + errorMsg;
}
// Get the result as a string
std::string output = lua["tostring"](result.get<sol::object>());
// Return the output
return output;
}
...
if (m_luaEditorData.empty())
m_luaEditorData.push_back('\0');
auto luaEditorFlags = ImGuiInputTextFlags_AllowTabInput | ImGuiInputTextFlags_CallbackResize;
ImGui::InputTextMultiline("##LuaEditor", m_luaEditorData.data(), m_luaEditorData.size(), ImVec2(ImGui::GetWindowContentRegionWidth(), ImGui::GetWindowHeight() - (ImGui::GetTextLineHeight() * 16)), luaEditorFlags, ResizeInputTextCallback, &m_luaEditorData);
...
When I run this code, I only get nil in my output, the correct output to stdout (don't really want it to output here, but to my std::string and when I put in bad code, it throws an exception in sol.hpp. I didn't really find any examples on how I can do this and I'm therefore am trying to figure this out on my own.
答案1
得分: 1
我通过用自己的print函数替换原来的print函数来解决了这个问题，如下所示：
int print(lua_State* L)
{
std::string printString = lua_tostring(L, 1);
g_luaOutput += printString + "\n";
return 1;
}
std::string Proxy::ExecuteLua()
{
std::string luaCode(m_luaEditorData.data());
sol::state lua;
lua.open_libraries(sol::lib::base, sol::lib::package, sol::lib::string, sol::lib::table, sol::lib::jit);
lua.set_function("print", print);
sol::protected_function_result result = lua.script(luaCode);
...
}
现在，当我在我的Lua编辑器中调用print函数时，它将不再打印到标准输出(stdout)，而是打印到我的自定义控制台上。
英文:
I was able to solve the problem myself by replacing the print function with my own like so;
int print(lua_State* L)
{
std::string printString = lua_tostring(L, 1);
g_luaOutput += printString + "\n";
return 1;
}
std::string Proxy::ExecuteLua()
{
std::string luaCode(m_luaEditorData.data());
sol::state lua;
lua.open_libraries(sol::lib::base, sol::lib::package, sol::lib::string, sol::lib::table, sol::lib::jit);
lua.set_function("print", print);
sol::protected_function_result result = lua.script(luaCode);
...
When I now call print in my lua editor I will not print to my own console instead of stdout.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论