• clayasaurus <clayasaurus gmail.com> Mar 24 2005
• "Ben Hinkle" <ben.hinkle gmail.com> Mar 25 2005
• clayasaurus <clayasaurus gmail.com> Mar 25 2005
• "Regan Heath" <regan netwin.co.nz> Mar 25 2005

clayasaurus <clayasaurus gmail.com> writes:

Hi all. I'm having trouble converting a couple of macros to a C or D 
function, I guess my problem is I don't know exactly what types are 
being passed to the macro or what it is really doing. Here is one of the 
macro's

#define lua_boxpointer(L,u) (*(void **)(lua_newuserdata(L, sizeof(void 
*))) = (u))

I know that L is of type lua_State*L, but I don't know what U is 
supposed to be, or why you can do strange things with it like set it 
equal to a function in a weird way. I think it is some sort of changable 
data (it could be different types). Anyway, I've tried to convert it to 
the following C function...

extern(C) {
void lua_boxpointer(lua_State *L, void ** u)
{
    *(void**)lua_newuserdata(L, sizeof(void*)) = u; // line 3
}

}

and I get a million errors

lua.d(3): found '*' when expecting '.' following 'void'
lua.d(3): found ')' when expecting identifier following 'void.'
lua.d(3): found ';' when expecting ','
lua.d(4): expression expected, not '}'

I'm not sure if there is a better way to figure out how to get these 
things implemented.

Anyway, any help would be greatly appreciated!

"Ben Hinkle" <ben.hinkle gmail.com> writes:

 extern(C) {
 void lua_boxpointer(lua_State *L, void ** u)
 {
    *(void**)lua_newuserdata(L, sizeof(void*)) = u; // line 3
 }


try
    *cast(void**)lua_newuserdata(L, (void*).sizeof) = u; // line 3

clayasaurus <clayasaurus gmail.com> writes:

Ben Hinkle wrote:
extern(C) {
void lua_boxpointer(lua_State *L, void ** u)
{
   *(void**)lua_newuserdata(L, sizeof(void*)) = u; // line 3
}


 
 try
     *cast(void**)lua_newuserdata(L, (void*).sizeof) = u; // line 3
 
 


Thanks, that compiled : ) And hopefully works too : )

"Regan Heath" <regan netwin.co.nz> writes:

On Fri, 25 Mar 2005 00:52:14 -0500, clayasaurus <clayasaurus gmail.com>  
wrote:
 Hi all. I'm having trouble converting a couple of macros to a C or D  
 function, I guess my problem is I don't know exactly what types are  
 being passed to the macro or what it is really doing. Here is one of the  
 macro's

 #define lua_boxpointer(L,u) (*(void **)(lua_newuserdata(L, sizeof(void  
 *))) = (u))

 I know that L is of type lua_State*L, but I don't know what U is  
 supposed to be, or why you can do strange things with it like set it  
 equal to a function in a weird way. I think it is some sort of changable  
 data (it could be different types). Anyway, I've tried to convert it to  
 the following C function...

 extern(C) {
 void lua_boxpointer(lua_State *L, void ** u)
 {
     *(void**)lua_newuserdata(L, sizeof(void*)) = u; // line 3
 }

 }

 and I get a million errors

 lua.d(3): found '*' when expecting '.' following 'void'
 lua.d(3): found ')' when expecting identifier following 'void.'
 lua.d(3): found ';' when expecting ','
 lua.d(4): expression expected, not '}'

 I'm not sure if there is a better way to figure out how to get these  
 things implemented.

 Anyway, any help would be greatly appreciated!


I'm not sure if this helps at all..  are you supposed to use C style or D  
style casts in an extern(C) block?

Regan