Pointer Incrementing Warning

Yaswanth,
if you use:

main()
{
   char xdata *Data_Ptr = 0x0000; /* Pointer to the 0x000 location */

   while(some loop)
   Data_Ptr++; /* Points to the next data location on the RAM-- here 0x0001 */
}

   char xdata *Data_Ptr = 0x0000; 

   char xdata *Data_Ptr;
   Data_Ptr = 0x0000; 

"you get same result"

NO you don't!

Data_Ptr++; // Advances the pointer to the next location

*Data_Ptr++ // Increments the value of the data pointed-to by the pointer!

I'm afraid, that's not true. The two statements are equivalent.
The ++ and * operators have same precedence but are evaluated right to left.
- Mike

Oh yeah.

Maybe that's the reason for the warning: the pointer has been incremented, and then dereferenced, but nothing has been done with the dereferenced value?

Vaclav,
I guess we need to do it this way.

char xdata *Data_Ptr;
Data_Ptr = (char *) 0x0000; /* I guess this is the way it must be done */

/* If done this way */
Data_Ptr = 0x0000;
/* you are not type casting it then in that case the compiler should/will give a warning -- Haven't tried it though. */

Data_Ptr++; /* Will point to the next location */
*Data_Ptr++; /* But surely this will point to the next location */

*Data_Ptr++;

 Proudly wasting time since 1981 

Andrew,

  *Data_Ptr++ // Increments the value of the data pointed-to by the pointer!

 (*Data_Ptr)++;

Hmm... Not sure what "the correct way" means. But the rest of the world increments a pointer this way:

Data_Ptr++;

++Data_Ptr;
or
Data_Ptr += 1;
or
Data_Ptr = Data_Ptr + 1;

*Data_Ptr++;
this expression is same as
*(Data_Ptr++);
or
Data_Ptr += 1, *Data_Ptr;

Got you...
Thanks Mike.

-Yaswanth

 Proudly wasting time since 1981 

Yaswanth,

" I guess we need to do it this way.

char xdata *Data_Ptr;
Data_Ptr = (char *) 0x0000; /* I guess this is the way it must be done */

/* If done this way */
Data_Ptr = 0x0000;
/* you are not type casting it then in that case the compiler should/will give a warning -- Haven't tried it though. */
"

There isn't necessary to do casting because it is only initialized pointer to char in xdata memory at location 0.
Compiler cannot be confused anyway I think. C51 Keil compiler don't generate warning in this case.

Rest of your question:
I agree with Mike's last and Andrew's second contribution.

Vaclav

I don't think the warning message is about pointer incrementing, instead, it is telling you that you are not doing any assignment operations in that loop - "expression with possibly no effect ".

I don't think the warning message is about pointer incrementing, instead, it is telling you that you are not doing any assignment operations in that loop - "expression with possibly no effect ".

That's exactly what it's telling you. Look folks, we need to learn ISO/ANSI C before we can start attacking the Keil compiler. If * didn't have precedence over ++ then how would the classic memcpy() example ever work? E.g.

while (count--)
{
    *pDest++ = *pSrc++;
}

int main(void)
{
    int status;
    for (;;) status; /* legal but useless */

    return 0;
}