Calling cint script from application

ROOT
1

Good time of the day to everyone.

I have a question.

There is a sample: C:\cint\demo\Win32App\graph01\

this sample demonstrates how to call external cint script from windows application. Here is the token of the code from that sample

	    char tmp[200];
	    PaintDC=BeginPaint(hWnd,&ps);
	    sprintf(tmp,"DrawGr((HDC)%ld",PaintDC);
	    G__calc(tmp); /* Call Cint parser */

it shows how to pass argement to the function.

my question is: (or rather to say questions are)

  1. AFAIU if I’m passing the ‘string’ argument the syntax will look like this:
	    sprintf(tmp,"MyCintFunction((const char*)\"%s\"","my string argument");

it works

BUT what if I’m have a structure.

for instance:

typedef struct SMyStructure
{
   int m_nVal;
   char m_sString[16];
} SMyStructure;

and fnuction which accept this structure and pointer to this structure as arguments, how syntax should look in this case?

thank you.

2

moreover
Am I in the right, saying cint scipt can’t modify any variables passed to it lilke pointers, or references?

I mean all arguments passed to cint script from application might be treated only like [in] variables. There are no any [out]s.

If I’m in the wrong, would you mind to point out how these varibales to be used.

thank you.

3

Assuming that CINT has the dictionary for SMyStructure and
that you have a function MyFunc(SMyStructure&):

SMyStructure myobj; sprintf(tmp,"MyFunc( * ((SMyStructure*)0x%x) )",&myobj); G__calc(tmp); /* Call Cint parser */
Cheers,
Philippe.

4

Yes, it should help.
I done something similar to strings

char* strMyString = "sdadas";
sprintf(tmp,"MyFunc( * ((void*)0x%x) )",(int)strMyString);
G__calc(tmp); /* Call Cint parser */

it works.
it is possible to change the context of the string also.

BUT whether it possible to put entire structure at the stack.

for instance I have a function in c script:

int fn_my_function(SMyStructure sMyStruct)
{
 ....
}

so, this is not a pointer, this is a structure itself.
waht would be syntax for calling then?

thank you.

5

The exact same syntax as the one when the function takes a reference.
Philippe.