Re: What's the best way to determine if a string is a reserved word in a C module?
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- Subject: Re: What's the best way to determine if a string is a reserved word in a C module?
- From: Tom N Harris <telliamed@...>
- Date: 2020年3月31日 06:45:43 -0400
On 3/31/2020 5:10 AM, Luyu Huang wrote:
I mean Lua reserved words. In fact, my C module accepts strings from Lua
and I need to determine if they are reserved words; they are essentially
TString objects. So if I can get the TString object, I can determine it
directly, Instead of comparing them with a reserved words list, which is
costly. I think it might be better for Lua to provide a function such as
`luaL_isreserved(lua_State *L, int idx)`.
Since a keyword can't be used as the name of a variable, the following
will determine if a word (at stack index) is reserved:
int status = LUA_OK;
int len = 0;
const char* str2 = NULL;
const char* str = lua_tolstring(L, index, &len);
// test if string matches the pattern "[%a_]%w*"
// left as exercise for the reader
if (is_lexical_word(len, str)) {
lua_pushvalue(L, index);
lua_pushstring(L, "=true");
lua_concat(L, 2);
str2 = lua_tolstring(L, -1, &len);
status = luaL_loadbuffer(L, str2, len, "=");
lua_pop(L, 2);
}
return status == LUA_ERRSYNTAX;
Attempting to load a string such as "function=true" will cause a syntax
error. You still need to check that the string can be a valid name or
else the string "black&tan" would be a syntax error even though it's not
a keyword.
However, I can't imagine running an entire Lua parser just to check one
string is a great use of resources when you could scan a list of
strings. Even if you don't want to repeat strcmp so many times, a clever
use of strstr will cut down the number of compares at the cost of
readability.
-- tom