Setting a pointerred value

hmnrobots · Joined: 02 Oct 2009 Posts: 11

Hi all
A 2 days Nightmare !
Chip is a 16f767, compiler 4.104
I declare a GLOBAL pointer:
int8 *FwdDev=0;
and a variable:
int8 d;

later then when I do:

(*FwdDev) = d;
Fprintf(HF,"\n\r RH d:%u %u Adr %u \n\r",d,*FwdDev,&FwdDev);

then I got:

RH d:177 0 Adr 80

That confirm the value of d is 177 but *FwdDev still remains null.I just don't understand!
any help please.

I even added:

*FwdDev = 5;
if(*FwdDev !=5) fprintf(HF,"\n\r No \n\r");

and I got always the answer NO
what is my mistake?
I added also #device *=16 : No success as well

sorry to disturb you with such a basic question

Mike Walne · Joined: 19 Feb 2004 Posts: 1785 Location: Boston Spa UK

Post the shortest possible complete compilable code which shows the problem you're having.
Then we can all see exactly what you are doing.

Mike

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

The pointer needs to point to some legitmate, allocated storage location
in RAM. Your code doesn't do that.

See the program below. One byte of RAM is allocated to the variable
'myval'. This allocation is done by the "int8 myval;" statement.
Now I can set the pointer, FwdDev, to point to myval. I do that by
taking the address of myval and putting it into FwdDev. Now it all
works. It displays 177.

The key thing you need to remember is that just simply creating a
pointer doesn't create or allocate any RAM storage locations for data.
You have a pointer, but you don't have anything for it to point to.
So to make the pointer be useful, you need to declare a variable, or
an array, and load the pointer with the address of that variable or array.

hmnrobots · Joined: 02 Oct 2009 Posts: 11

JUST THANK YOU SO MUCH , at least I've got an answer and I will remember that declaring a pointer doesn't allocate any memory!

Once again Many thanks PCM Programmer to take the time to answer to such a basic question.