lua-users home
lua-l archive

[Date Prev][Date Next][Thread Prev][Thread Next] [Date Index] [Thread Index]

Ok, well, what i'd like to do is this:

lua_Object lo = tolua_getobject( ParamOne, 0 );
and assuming Param One is a string -

I'd like to check the strlen and get the data if 
there is any:

size  = lua_strlen( lo );
lstr  = lua_getstring( lo );

from here, i can recreate a new string with the contents of
lstr plus any data I wish to add to it. (say lstr is at
first an empty lstring.)

But I can't seem to get this data back into lstr, so 
that a call like this would return the new buffer in the
old parameter.  

success/failure func( buffer, size )

My biggest concern here is being fast, I'd rather
not invoke any tag methods or have to create new strings
or anything.

I'd just like a simple:

lua_expandlstring( lo, newbuffersize ) I can get the lstr buffer pointer and copy new (sometimes
larger) data chunks into it.


-----Original Message-----
From: Luiz Henrique de Figueiredo []
Sent: Wednesday, May 31, 2000 5:35 AM
To: Multiple recipients of list
Subject: Re: lstring expansion from the c side

>From  Tue May 30 15:44:16 2000
>From: Jim Mathies <>
>What is the simplest and fastest way to
>expand the size of the buffer used to
>store a lua lstring, from the c side?
>assuming I have the string object.

I'm sorry, I don't quite understand what you want to do.
There's no way to change individual bytes inside a Lua string -- you
have to
create another string.
So, if you want to edit a string, just do it in C (say, by changing
individual bytes or by appending something to it), and then send this
to Lua, using lua_pushlstring.
Perhaps I'm missing something.