//c++ funciton definition
int LS_SomeFucntion(LuaState* state)
{
LuaStack args(state);
//..
set<int>::iterator it = mySet.begin();
for(; it != mySet.end(); ++it)
{
state.pushInteger(*it);
}
return mySet.size();
}
state->GetGlobals().Register("SomeFunction",LS_SomeFunction);
//lua scripts
??? = SomeFunction()
如何在lua脚本中获取SomeFunction()的返回值 何时调用函数时不知道大小?
答案 0 :(得分:4)
您可以捕获表格中的所有返回值:
local rv = { SomeFunction() }
print('SomeFunction returned', #rv, 'values')
for i,val in ipairs(rv) do
print(i,val)
或者使用变量参数列表处理它们:
function DoSomething(...)
local nargs = select('#', ...)
print('Received', nargs, 'arguments')
for i=1,nargs do
print(i,select(i,...))
end
DoSomething(SomeFunction())
当然,你的C函数应该只返回一个包含列表项的Lua表。我不熟悉LuaPlus,但从文档here来看,你需要这样的东西:
int LS_SomeFunction(LuaState* state)
{
LuaObject table;
table.AssignNewTable(state, mySet.size()); // presize the array portion
int i = 1;
for(set<int>::iterator it = mySet.begin(); it != mySet.end(); ++it)
table.SetNumber(i++, *it);
table.PushStack();
return 1;
}
然后你就说:
local rv = SomeFunction()
print('SomeFunction returned', #rv, 'values')