使用 Lua C API 迭代 table 的 table 秒

Question

我正在尝试迭代 Lua 中的 table 个 table 并输出：

每个table的键。
每个table中每个条目的键/值对。

代码如下：

void print_table(lua_State *L)
{
    lua_pushnil(L);
    while(lua_next(L, -2) != 0) {
    const char *key = lua_tostring(L, -2);
            if(lua_isstring(L, -1))
                printf("%s = %s", key, lua_tostring(L, -1));
            else if(lua_isnumber(L, -1))
                printf("%s = %d", key, lua_tonumber(L, -1));
            else if(lua_istable(L, -1)) {
                printf("%s", key);
                PrintTable(L);
            }
            lua_pop(L, 1);
        }
    }
}

这是我尝试输出的 table 之一的示例：

s = {
        p = {
            n = "D",
            g = "1",
        },
        d = {
            l = "N",
            p = "N",
            u = "O",
            po = 100,
        },
        e = {
            {
                n = "B",
                l = "P",
                p = "P",
                u = "P",
                po = "P",
                pa = {
                    v = "4",
                    a = "U",
                    f = {
                        { name = "U", type = "U" },
                        { name = "A", type = "I" },
                        { name = "A", type = "I" },
                        { name = "P", type = "U" },
                        { name = "P", type = "U" },
                        { name = "P", type = "I" },
                        { name = "T", type = "U" },
                        { name = "D", type = "U" },
                        { name = "D", type = "I" },
                        { name = "S", type = "I" },
                        { name = "C", type = "U" },
                        { name = "G", type = "U" },
                        { name = "C", type = "F" },
                        { name = "C", type = "U" },
                    },
                },
                c = {
                    v = "1",
                    a = "",
                    f = {
                        { name = "B", type = "U" },
                        { name = "E", type = "F" },
                    },
                },
            },
        },
    }

函数在以下行崩溃：

while(lua_next(L, -2) != 0)

由于索引无效。导致崩溃的脚本行是：

{ name = "B", type = "U" },

我不得不承认我对 Lua 中的堆栈不是很熟悉，我尝试搜索类似的答案但找不到任何答案。有人知道我做错了什么吗？

谢谢！

添加工作版本以防有人感兴趣：

void print_table(lua_State *L)
    {
        if ((lua_type(L, -2) == LUA_TSTRING))
            printf("%s", lua_tostring(L, -2));

        lua_pushnil(L);
        while(lua_next(L, -2) != 0) {
            if(lua_isstring(L, -1))
                printf("%s = %s", lua_tostring(L, -2), lua_tostring(L, -1));
            else if(lua_isnumber(L, -1))
                printf("%s = %d", lua_tostring(L, -2), lua_tonumber(L, -1));
            else if(lua_istable(L, -1)) {
                print_table(L);
            }
            lua_pop(L, 1);
        }
    }

Answer 1

f = {
    { name = "B", type = "U" },
    { name = "E", type = "F" },
}

相当于：

f = {
    [1] = { name = "B", type = "U" },
    [2] = { name = "E", type = "F" },
}

当您在键上调用 lua_tostring 时 Lua 会将数字索引更改为字符串。

const char *key = lua_tostring(L, -2);

lua_tostring 使用 lua_tolstring 并从 manual:

If the value is a number, then lua_tolstring also changes the actual value in the stack to a string. (This change confuses lua_next when lua_tolstring is applied to keys during a table traversal.)

最好使用 lua_type 检查键是否真的是字符串，因为 lua_isstring 只会告诉你堆栈值是否可以转换为字符串。您还可以推送密钥的副本并在副本上调用 lua_tostring。

使用 Lua C API 迭代 table 的 table 秒

Iterating over table of tables with the Lua C API

c

lua

lua-table

lua-api