如何使用作为参数传递给一个lua C函数传递表工作? [英] How to work with tables passed as an argument to a lua C function?
问题描述
我要实现与C语言的函数,将由Lua的脚本调用。
这个功能应该得到一个Lua表作为参数,所以我应该读取table.I领域尝试做如下图所示,但是当我运行我的功能崩溃。谁能帮我找到了问题?
/ *
功能findImage(选项)
的ImagePath = options.imagePath
模糊= options.fuzzy
ignoreColor = options.ignoreColor;
结束 调用示例: findImage {=的ImagePath的/ var / image.png,模糊= 0.5,ignoreColor = 0XFFFFFF} * /
//实现用C语言函数
静态INT findImgProxy(lua_State * L)
{
luaL_checktype(L,1,LUA_TTABLE); lua_getfield(L,-1,的ImagePath);
如果(!lua_isstring(L,-1)){
错误();
}
为const char *的ImagePath = lua_tostring(L,-2);
调用lua_pop(L,1); lua_getfield(L,-1,模糊);
如果(!lua_isnumber(L,-1)){
错误();
}
浮模糊= lua_tonumber(L,-2); lua_getfield(L,-1,ignoreColor);
如果(!lua_isnumber(L,-2)){
错误();
}
浮ignoreColor = lua_tonumber(L,-2); ... 返回1;
}
如何从C返回表到Lua:的
结构点{
INT X,Y;
}
的typedef点点;
静态INT returnImageProxy(lua_State * L)
{
点点[3] = {{11,12},{21,22},{31,32}}; lua_newtable(L); 的for(int i = 0;我3;我++){
lua_newtable(L);
调用lua_pushnumber(L,点[I] .X);
lua_rawseti(L,-2,0);
调用lua_pushnumber(L,点[I] .Y);
lua_rawseti(L,-2,1);
lua_settable(L,-3);
} 返回1; //我要回一个Lua表所示:{{11,12},{21,22},{31,32}}
}
在使用Lua的C API工作它来获得舒适的虚拟堆叠的工作是很重要的 - 所有重要的语言边界的相互作用发生在那里。看着你code片段,它看起来并不像你在正确的编组数据为C。
在写一个lua的C函数,你基本上做3件事情:
- 转换输入数据卢阿到的东西,你可以在C一起工作。
- 执行处理或任何功能需要做。
- 转换并返回如有回卢阿输出结果。
作为一个例子,这里是你的 findImgProxy
应该是这样的:
静态INT findImgProxy(lua_State * L)
{
//丢弃传入任何额外的参数
lua_settop(L,1);
luaL_checktype(L,1,LUA_TTABLE); //我们获取数据的表
//'解包'通过将值搬上了台面
//堆栈第一。然后转换这些堆栈值
//为一个适当的C型。
lua_getfield(L,1的ImagePath);
lua_getfield(L,1,模糊);
lua_getfield(L,1,ignoreColor);
//堆栈现在有以下几点:
// 1 = {=的ImagePath的/ var / image.png,模糊= 0.5,ignoreColor = 0XFFFFFF}
// -3 =/var/image.png
// -2 = 0.5
// -1 = 0XFFFFFF 为const char *的ImagePath = luaL_checkstring(L,-3);
双模糊= luaL_checknumber(L,-2);
INT ignoreColor = luaL_checkint(L,-1);
//我们可以弹出模糊ignoreColor从堆栈
//因为我们通过价值得到了他们
调用lua_pop(L,2); //做处理功能
// ... 返回1;
}
请注意,我们必须保持的ImagePath
在堆栈上,因为我们持有为const char *
它。弹出该字符串将关闭无效 *的ImagePath
自卢阿可以收集它。
另外,你可以通过复制 luaL_checkstring
返回到另一个缓冲区中的字符串。在这种情况下,突然出现的串关是确定的,因为我们不再指向由LUA拥有内部缓冲区。
编辑:如果有的在表中的键都是可选的,你可以使用 luaL_opt *
功能,而不是提供的默认值。例如,如果模糊
和 ignoreColor
是可选的:
// ...
为const char *的ImagePath = luaL_checkstring(L,-3);
双模糊= luaL_optnumber(L,-2,0.0); //默认为0.0,如果没有模糊
INT ignoreColor = luaL_optint(L,-1,0); //默认为0,如果没有ignoreColor
// ...
因此,如果调用code提供了一个关键的一个无意义的值,这仍然会产生一个错误。 OTOH,如果它不存在则该值为无
并提供了使用默认值。
I'm going to implement a function with C language and which will be called by Lua script.
This function should receive a lua table as the argument, so I should read the fields in the table.I try to do like below, but my function is crashing when I run it. Can anyone help my find the problem?
/*
function findImage(options)
imagePath = options.imagePath
fuzzy = options.fuzzy
ignoreColor = options.ignoreColor;
end
Call Example:
findImage {imagePath="/var/image.png", fuzzy=0.5, ignoreColor=0xffffff}
*/
// implement the function by C language
static int findImgProxy(lua_State *L)
{
luaL_checktype(L, 1, LUA_TTABLE);
lua_getfield(L, -1, "imagePath");
if (!lua_isstring(L, -1)) {
error();
}
const char * imagePath = lua_tostring(L, -2);
lua_pop(L, 1);
lua_getfield(L, -1, "fuzzy");
if (!lua_isnumber(L, -1)) {
error();
}
float fuzzy = lua_tonumber(L, -2);
lua_getfield(L, -1, "ignoreColor");
if (!lua_isnumber(L, -2)) {
error();
}
float ignoreColor = lua_tonumber(L, -2);
...
return 1;
}
How about return a table from C to Lua:
struct Point {
int x, y;
}
typedef Point Point;
static int returnImageProxy(lua_State *L)
{
Point points[3] = {{11, 12}, {21, 22}, {31, 32}};
lua_newtable(L);
for (int i = 0; i 3; i++) {
lua_newtable(L);
lua_pushnumber(L, points[i].x);
lua_rawseti(L, -2, 0);
lua_pushnumber(L, points[i].y);
lua_rawseti(L, -2, 1);
lua_settable(L,-3);
}
return 1; // I want to return a Lua table like :{{11, 12}, {21, 22}, {31, 32}}
}
When working with the Lua C API it's important to get comfortable working with the virtual stack -- all the important language boundary interactions happen there. Looking at your code snippet, it does not look like you're marshaling the data properly over to C.
When writing a lua C function you basically have to do 3 things:
- Convert input lua data into something you can work with in C.
- Perform the processing or whatever the function needs to do.
- Convert and return the output result if any back to lua.
As an example, here's what your findImgProxy
should look like:
static int findImgProxy(lua_State *L)
{
// discard any extra arguments passed in
lua_settop(L, 1);
luaL_checktype(L, 1, LUA_TTABLE);
// Now to get the data out of the table
// 'unpack' the table by putting the values onto
// the stack first. Then convert those stack values
// into an appropriate C type.
lua_getfield(L, 1, "imagePath");
lua_getfield(L, 1, "fuzzy");
lua_getfield(L, 1, "ignoreColor");
// stack now has following:
// 1 = {imagePath="/var/image.png", fuzzy=0.5, ignoreColor=0xffffff}
// -3 = "/var/image.png"
// -2 = 0.5
// -1 = 0xffffff
const char *imagePath = luaL_checkstring(L, -3);
double fuzzy = luaL_checknumber(L, -2);
int ignoreColor = luaL_checkint(L, -1);
// we can pop fuzzy and ignoreColor off the stack
// since we got them by value
lua_pop(L, 2);
// do function processing
// ...
return 1;
}
Note that we must keep imagePath
on the stack since we're holding a const char *
to it. Popping that string off would invalidate *imagePath
since lua might collect it.
Alternatively, you can copy the string returned by luaL_checkstring
into another buffer. Popping the string off in this case is ok since we're no longer pointing to an internal buffer owned by lua.
Edit: If some of the keys in the table are optional, you can use the luaL_opt*
functions instead and provide defaults. For example, if fuzzy
and ignoreColor
are optional:
// ...
const char *imagePath = luaL_checkstring(L, -3);
double fuzzy = luaL_optnumber(L, -2, 0.0); // defaults to 0.0 if no fuzzy
int ignoreColor = luaL_optint(L, -1, 0); // defaults to 0 if no ignoreColor
// ...
So if the calling code provides a nonsensical value for a key, this will still raise an error. OTOH, if it's absent then the value is nil
and the default provided is used instead.
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