Question

I have strange problem, with calling Lua function from C++. I have in Lua:

Player = 
{
    Number = 0.43,
    Text = "SomeText",
}

function Player:Func(a, b)  
    return (a * b);
end

Before lua_pcall my stack looks:
table
function
3
4

I call this function with:

lua_pcall(L, 2, 1, 0)

我从Lua那里得到错误:

尝试对当地b(无价值)进行算术

当我改变卢阿的脚本

return (a * b);

至

return a;

There is no error, but from lua_至number(L, -1); I get value 4 (my second argument in C:/), so it looks that my second argument in C is first in Lua. Do you know what I made wrong in my code ?
How I construct stack:

lua_getglobal (L, "Player");
lua_pushstring(L, "Func");
lua_gettable(L, -2);
lua_pushnumber(L, 3.0);
lua_pushnumber(L, 4.0);

Answer 1

Bens评论是关键 - 阅读“Lua的方案编制”第150页中面向目标的方案拟订部分。

http://www.lua.org/pil/16.html>http://www.lua.org/pil/16.html

The effect of the colon is to add an extra hidden parameter in a method definition and to add an extra argument in a method call.

所以您需要将“ 账户” 对象作为第一个参数, 或者( 在此情况下更容易) 将 < code> 函数玩家: func( a, b) 改为 < code > 函数玩家。 func( a, b)

友情链接