Thread: Problems returning char *

Hi!, I'm new and speak spanish...Therefore I hope that you can understand me and help.

My problem is the next:

I have a function with this statement

char * consultaCuenta(char *buffer, char *status)

Inside I declare a variable:

char retorno[199];

and then i do this (after fill retorno with data):

sprintf(buffer,"%189.189s",retorno)

return (buffer);

I receive the return in a variable:

char *v;

The compiler says (I translate):

"the return make (build??) a pointer from an integer without a conversion"

how I did wrong? what's the problem?

thanks for the help.

Sorry...here the code goes :-)

The source is not pure C is Pro*C (but the pointers are the same :-) )

Code:

/* Global variables */
EXEC SQL BEGIN DECLARE SECTION;
    retorno[199];
EXEC SQL END DECLARE SECTION;


char * consultaCuenta(char *buffer)
{
      EXEC SQL WHENEVER SQLERROR DO      return (sqlError(buffer));
      EXEC SQL DECLARE CD CURSOR for
        select lpad(num_doc,11,' ')|  lpad(MONTO,11,' ')||lpad(NUM_CONSEJERA,12,' ')||estado
           from AVON_CAJAVEC
           where
           num_doc = lpad(:pecnmro_servicio,10,'0');

         EXEC SQL OPEN CD;
         EXEC SQL WHENEVER NOT FOUND DO   break;

           for (;;)
             {
               EXEC SQL FETCH CD INTO :retorno;
               sprintf(buffer,"%198.198s",  retorno);
             }
         EXEC SQL CLOSE CD;
 return (buffer);
}

the function where the results is received is too long to put here, but the important sentences are this:

char *buff_aux;

 buff_aux = consultaCuenta(buffer);

thanks again

Pro*c is the same than C but with SQL sentences...nothing more.

Yes!!...the return make (build??) a pointer from an integer without a conversion...is in that line the warning.

I'm sorry...i'm wrong with the declaration.

The type is char...i forgot to put there...

char retorno[199]

that's the declaration...

Thanks Elysia for your time and patience. The mistake must be there in other place. I'll check the entire code...thanks

Greetings from Chile!!

I'm not sure if this is the cause of your problem, but one bad thing I see is that you pass a char* to the function without passing the size of the char* array, then you write to it. If you pass a pointer to a function that is going to be written to, you should always pass the size of the buffer, otherwise you might write too much data.