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russk2txb
Joined: 30 Mar 2008
Posts: 109
Location: New Jersey
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| 8 bit structure address? |
 Posted: Wed Mar 09, 2016 10:00 am |
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Here is a small test program to show my problem | Code: | typedef struct {
int index;
int16 addr;
int location;
int dtype;
} TableAddrInfo;
TableAddrInfo tio = { 1,2,3,4 };
void main ()
{
TableAddrInfo *pTi;
void *p;
p = &tio;
pTi = &tio;
while (1);
}
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The chip is 18F6723. The compiler is version 5.053. The problem is that the compiler thinks that pTi is an 8 bit pointer. Setting a breakpoint at the while statement, the value of p is 28B. The value of pTi is 8B. Why?
Thanks, Russ |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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| Posted: Wed Mar 09, 2016 11:49 am |
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| russk2txb wrote: |
The compiler thinks that pTi is an 8 bit pointer. Setting a breakpoint at the
while statement, the value of p is 28B. The value of pTi is 8B.
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I installed vs. 5.053 and compiled the following program. Then I ran it
in MPLAB vs. 8.92 simulator. The output is shown below. Conclusion:
the compiler is using 16-bit pointers for the structure address.
| Quote: |
p = 028b
pTi = 028b
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| Code: | #include <18F6723.h>
#fuses INTRC_IO,NOWDT,PUT,BROWNOUT,NOLVP
#use delay(clock=4M)
#use rs232(baud=9600, UART1, ERRORS)
typedef struct {
int index;
int16 addr;
int location;
int dtype;
} TableAddrInfo;
TableAddrInfo tio = { 1,2,3,4 };
#locate tio = 0x28B
//========================
void main()
{
TableAddrInfo *pTi;
void *p;
p = &tio;
pTi = &tio;
printf("p = %lx \r", p);
printf("pTi = %lx \r", pTi);
while(TRUE);
} |
I don't have your full program with many ram variables, so I used the
#locate directive to set the address of 'tio' to 0x28B. Can you add code
to your program so it prints out the pointer values ? Don't use the
debugger with breakpoints and "mouse over". Just print them out with
"%lx". What do you get ? |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19520
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| Posted: Wed Mar 09, 2016 3:24 pm |
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and the PIC18, does not use 8bit pointers. My guess would be that whatever form of debugging you are using 'thinks' that pTi, is an 8bit variable.
As an update to this, a void pointer, will be a 16bit pointer, but until cast to another type, it'll be incremented in ones, while a pointer to another type is incremented in the size of this type. If you want to access the second element of a type, pointed to by a void pointer, you have to cast it to the type _before_ incrementing. |
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russk2txb
Joined: 30 Mar 2008
Posts: 109
Location: New Jersey
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| Posted: Thu Mar 10, 2016 3:50 am |
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Thank you PCM. The ^&*(&*() debugger does it to me again! Using a printf as you did does indeed show the correct address. But if you use the debugger to show the address the pTi value will be shown truncated. Try hovering over the values with the mouse in the debugger, or using the eval window, and see if you get the same result, please.
When having problems with a piece of code the debugger has always been my second line of attack when I cannot see the problem by inspection of the code. But I am constantly having to learn how and when it fails in some way...
Regards, Russ |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19520
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| Posted: Thu Mar 10, 2016 4:46 am |
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In MPLAB, when you create a watch, you have the option to specify the size of the object being watched.
What is happening, is that MPLAB, is pulling from the Coff data the fact that the pointer points 'to' an 8bit value, and incorrectly defaulting to displaying the pointer 'as' an 8bit value. Manually specify the size as 16bit, and it should work fine. |
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