我无法用一个简短的描述性标题来解释我的问题,对不起。
我正在从c ++调用一个名为hook.Call(event,...)的lua函数。它调用hook.Add(event,unique_name,function)添加的所有函数
问题是当我在一个钩子里面调用print(...)函数时,它不会打印出你希望它打印的东西,因为调用一个钩子的堆栈仍然存在。所以它从那个堆栈打印出来。 我无法删除堆栈,因为那样我就无法从钩子中获取返回值。
钩子调用如下所示:
int CGame::Update(bool haveFocus, unsigned int updateFlags)
{
hook::StartCall("PreGameUpdate");
hook::PushInteger(haveFocus);
hook::PushInteger(updateFlags);
hook::CallReturn(1, 2); //Calls hook.Call("PreGameUpdate", haveFocus, updateFlags) here
//The first argument is the amount of values to return to C++
//The second argument is how many values that were pushed onto the stack (maybe I can monitor that?)
if(hook::IsBoolean(1) && hook::GetBoolean(1) == false)
return false; //skip the rest of the code if we return false in the Pre hook
hook::End();
//rest of the code in CGame::Update() here
}
我正在考虑打印下一帧,但听起来真的很糟糕,我甚至不确定我是怎么做的。
钩子功能
namespace hook
{
void StartCall(string hookname)
{ lua_State *L = LuaJIT::GetState();
//Remove any previous stack (just in case?)
//Stack is now: nil
lua_settop(L, 0);
//Get the "hook" global and check if it exists
//stack is now: hook
lua_getglobal(L, "hook");
if (!lua_istable(L, -1))
return;
//Get the function "Call" and check if it exists
//Stack is now: hook.Call()
lua_getfield(L, -1, "Call");
if (!lua_isfunction(L, -1))
return;
//Remove the "hook" table from the stack leaving only the Call function left
//Stack is now: Call()
lua_remove(L, 1);
//Push the hookname onto the stack
//Stack is now: Call(hookname)
lua_pushstring(L, hookname);
}
void CallReturn(int returns, int args)
{ lua_State *L = LuaJIT::GetState();
//PrintStack("PRE PCALL");
/* When printing the stack, this is the output:
===========PRE PCALL=================START
1:
function: 2116D588
2:
PreGameUpdate
3:
1.000000
4:
0.000000
===========PRE PCALL=================END
*/
//Check if the first value is a function and if the stack is valid
if (!lua_isfunction(L, 1) || lua_gettop(L) == 0)
{
PrintStack("NO FUNCTION IN STACK");
return;
}
//pcall "Call" from the stack and check if it worked
//Stack is now: pcall(Call(hookname, ...))
int status = lua_pcall(L, args + 1, returns, 0);
//Check if it errored
if(status != 0)
{
//Print to debug output if it errored
PrintStack("STACK");
Msg("PCALL ERROR[%s]: %s", lua_tostring(L, 1), lua_tostring(L, -1));
}
//PrintStack("POST PCALL");
}
void End()
{ lua_State *L = LuaJIT::GetState();
//Remove the stack again
//Stack is now: nil
lua_settop(L, 0);
}
void EndNoReturn(int args)
{
CallReturn(0, args);
End();
}
void StartCallNoPush(string hookname, int returns)
{
StartCall(hookname);
CallReturn(0, returns);
}
void CallSimple(string hookname)
{
StartCall(hookname);
CallReturn(0, 0);
End();
}
void PushBoolean(bool res)
{ lua_State *L = LuaJIT::GetState();
int test = toint(res);
lua_pushboolean(L, test);
}
bool GetBoolean(int idx)
{ lua_State *L = LuaJIT::GetState();
int res = lua_toboolean(L, idx);
lua_pop(L, 1);
return tobool(res);
}
int IsBoolean(int idx)
{ lua_State *L = LuaJIT::GetState();
int res = lua_isboolean(L, idx);
lua_pop(L, 1);
return res;
}
//The rest of the code is just like the three functions above but for different types
}
打印功能
int print(lua_State *L)
{
//PrintStack("PRINT PRE");
int n = lua_gettop(L); /* number of arguments */
int i;
lua_getglobal(L, "tostring");
for (i=1; i<=n; i++) {
const char *s;
lua_pushvalue(L, -1); /* function to be called */
lua_pushvalue(L, i); /* value to print */
lua_call(L, 1, 1);
s = lua_tostring(L, -1); /* get result */
if (s == NULL)
return luaL_error(L, LUA_QL("tostring") " must return a string to "
LUA_QL("print"));
if (i>1) CryLogAlways("\t");
CryLogAlways(s);
lua_pop(L, 1); /* pop result */
}
CryLogAlways("\n");
//PrintStack("PRINT POST");
return 0;
}
我没有完成大部分的打印功能。我从我朋友的代码中取出它,这就是为什么它没有像那些钩子函数那样被注释。当没有在钩子中调用时,打印确实有效。
因此print的问题在于它会打印挂钩堆栈中的所有内容,因为在删除堆栈之前它已被调用。
我还发现推送和弹出非常令人困惑,所以它真的有助于注释代码,就像在钩子调用函数中显示堆栈当前是什么。
我猜测整个问题是c ++中钩子函数的一个设计缺陷,但我真的不知道我是怎么做的。
答案 0 :(得分:1)
我在int print的底部弹出了堆栈的tostring作为注释中提到的interjay,它现在应该像它一样工作。
我很抱歉没有足够的描述性。